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3 years ago

WHERE ARE THE EXPERTS AND ACE!!!!!!! I NEED HELP PLS SHARE YO SMARTNESS!!!!! WILL GIVE BRAINLIEST AND RATE AND VOTE!!!

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

5 0

Answer:

Value of tan( π/3 ):

${ \tt{ \frac{\pi}{3} = \frac{180}{3} = 60 \degree }}$

60° lies within the first quadrant.

${ \sf{ \tan( \frac{ \pi}{3} ) = ( \frac{1}{2} , \: \frac{ \sqrt{3} }{2}) }}$

Since it's a half:

${ \boxed{ \sf{ \tan( \frac{\pi}{3} ) = \sqrt{3} }}}$

Length of hypotenuse:

${ \bf{ \sin(31 \degree) = \frac{16.5}{h} }} \\ { \sf{h = 32.04}}$

Hypotenuse = 32.0

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3 years ago

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Aneli [31]

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3 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

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

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

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

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3 years ago

Can somebody help me on this thank you it will mean a lot

MaRussiya [10]

Answer:

your will be A

Step-by-step explanation:

hoped i helped and can i get brainiest pls

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