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Fittoniya [83]
2 years ago
5

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
Read 2 more answers
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 />
4 0
3 years ago
Isabelle is mixing red paint with blue paint to make purple paint. She adds 3/10 of a fluid ounce of red to 11/15 of a fluid oun
MAXImum [283]
40\13 that what she needs
6 0
3 years ago
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If the sphere shown above has a radius of 14 units then what is the approximate volume of the sphere
oksano4ka [1.4K]

Answer:

About 11494.04

Step-by-step explanation:

There is no sphere shown above but I can still get you the answer:

The formula for the volume of a sphere is \frac{4}{3} *π*r³

Putting in the radius, we get:

\frac{4}{3}*π*14³=

\frac{4}{3}*π*2744=

\frac{10976}{3}*π=

3658.66*π≈11494.04

8 0
3 years ago
Krutika and Gavin have played 41 tennis matches.
Neko [114]

Answer:

P(Kritika)= 27/41

P(Gavin)=1-(27/41)

= (41-27)/41

=14/41

5 0
3 years ago
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