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raketka [301]
3 years ago
10

the cycle of processes by which water circulates between the earth's oceans, atmosphere, and land, involving precipitation as ra

in and snow, drainage in streams and rivers, and return to the atmosphere by evaporation and transpiration.
Mathematics
1 answer:
lina2011 [118]3 years ago
6 0
Yes they do because the sun evaporates the water and then it circles around and ends up in the sky
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Exponents: Can me help me? This is the one, I just can't get.
aliina [53]

Answer:

1

Step-by-step explanation:

x^y * x^-y

When the bases are the same,   we add the exponents when multiplying

x^(y-y)

x^0

Anything to the zero power is 1  (except 0)

1


4 0
3 years ago
the total cost for two items is $19 is the difference in the cost of the two items is $5 find the cost of each item.
lora16 [44]

Ok so you spent 19 dollars the difference between prices is 5 dollars and your answer would be 12 and 7. If you need any help let me know.

5 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
What is m∠LQK ?<br><br> Enter your answer in the box.
Lady_Fox [76]
∠LQK+∠GQL=90º
Therefore:
(4n-15)+(3n)=90
7n=90+15
7n=105
n=105/7
n=15

∠LQK=4n-15=4(15)-15=60-15=45

Answer: ∠LQK=45º
5 0
3 years ago
Read 2 more answers
Can someone please help me :) !!!!
Gwar [14]

Answer:

A

Step-by-step explanation:

supplementary angles add up to 180

angle cgr and kga are vertical angles so they are equal

that makes the two angles in question add up to 180

6 0
3 years ago
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