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melisa1 [442]
3 years ago
5

Drag the tiles to the boxes to form correct pairs

Mathematics
1 answer:
sergiy2304 [10]3 years ago
3 0

( -  \frac{3}{4} )( \frac{7}{8} )

=  \frac{ - 3}{4}  \times  \frac{7}{8}

=  \frac{ - 21}{32}

_______________________________________

( \frac{2}{3} )( - 4)(9)

=  \frac{2}{3}  \times  - 4 \times 9

=  \frac{ - 72}{3}

=  - 24

_______________________________________

( \frac{5}{16} )( - 2)( - 4)( -  \frac{4}{5} )

=  \frac{5}{16}  \times  - 2 \times  - 4 \times  \frac{ - 4}{5}

=  \frac{ - 160}{80}

=  \frac{ - 16}{8}

=  - 2

_______________________________________

(2 \frac{3}{5} )( \frac{7}{9} )

=  \frac{13}{5}  \times  \frac{7}{9}

=  \frac{91}{45}

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3 years ago
Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).
lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted C, which we can parameterize by the vector-valued function,

\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)

for 0\le t\le1, which has differential

\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

Then with x(t)=y(t)=z(t)=1-t, we have

\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r

=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

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siniylev [52]
The absolute value of rational number 3.7 is 3.7
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What is (-6,-2)×1/2 ?​
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Answer:

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