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3 years ago

AYÚDENME!!!! T.T PORFA

Mathematics

1 answer:

Vitek1552 [10]3 years ago

7 0

Answer:

Por el teorema de 30 grados en un triangulo rectangulo la linea roja es $\sqrt{3}$ veces la longitud del cateto opuesto al angulo de de 30. Por tanto la linea roja será:

$l=\sqrt{3}\times3\sqrt{3}=9$

Luego

$x=9-3=6$

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lesya [120]

Where is the question?????

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3 years ago

10 points to the person who answers this whole thing and I will mark you as brainliest.

olga2289 [7]

Answer:

Step-by-step explanation:

Perimeter is found by adding together all the lengths of the sides. For us, that is x + (2x + 5) + (6x - 17) + (3x + 2). Now we will just combine like terms. We can also drop the parenthesis because they do nothing for us and mean nothing to the problem.

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viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
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The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
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$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

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