Which diagram represents the hypothesis of the converse of corresponding angles theorem?

Mathematics

1 answer:

BabaBlast [244]2 years ago

5 0

Answer:

the first diagram

Step-by-step explanation:

first one

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Which set of ordered pairs could be generated by an exponential function? A)(negative 1, negative one-half), (0, 0), (1, one-hal

Vadim26 [7]

Answer:

<h2>C) (negative 1, one-half), (0, 1), (1, 2), (2, 4)</h2>

Step-by-step explanation:

The set C could be generated by an exponential function. The main reason is that exponential functions hava a restricted range, it can't have negative numbers or the number zero, because power can only be equal or greater than 1.

Additionally, for all exponentials, a null exponent gives 1 as an answer, so point (0, 1) is always present in an exponential function.

Therefore, the right answer is C.

3 0

3 years ago

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Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$ <span />