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Elza [17]
3 years ago
8

ddle" class="latex-formula">


Last one! bye!
Thanks!
Mathematics
2 answers:
tekilochka [14]3 years ago
8 0

Step-by-step explanation:

6 - 2x + 5 + 4x \\ 2x + 11 = 0 \\ 2x =  - 11 \\ x =  \frac{ - 11}{2}  \\ x =  - 5.5

hope helpful <3

wolverine [178]3 years ago
6 0

\sf 6-2x+5+4x

\sf \longmapsto - 2x + 4x + 6 + 5

\sf \longmapsto  - 2x + 4x + 11

\sf \longmapsto2x + 11

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attashe74 [19]
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2.6 x 2.8 = 7.28
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charle [14.2K]

Answer:

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Step-by-step explanation:

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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
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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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8 0
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