Answer:
So there are 500 pairs of numbers that have a sum of 1001. Thus, the sum of numbers from 1 to 1000 is 500*1001 = 500,500.
Step-by-step explanation:
Answer: X = 3t, Y =2 - t, Z =2
Step-by-step explanation: the plane
x + y + z =4has normal vector
M =<1,1,1> and the line
x = 1 + t, y = 2 − t, z = 2t has direction
v =<1, −1, 2>. So the vector
A= n × v
=<1, 1, 1> × <1, −1, 2>
=<2−(−1),1−2,−1−1>
=<3,−1,−2>
The range of the answer is [-3,infinity) and {yly>=-3}
Make a change of coordinates:
![u(x,y)=xy](https://tex.z-dn.net/?f=u%28x%2Cy%29%3Dxy)
![v(x,y)=\dfrac xy](https://tex.z-dn.net/?f=v%28x%2Cy%29%3D%5Cdfrac%20xy)
The Jacobian for this transformation is
![\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20J%3D%5Cbegin%7Bbmatrix%7D%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%26%5Cdfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D%5C%5C%5C%5C%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D%26%5Cdfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7Dy%26x%5C%5C%5C%5C%5Cdfrac1y%26-%5Cdfrac%20x%7By%5E2%7D%5Cend%7Bbmatrix%7D)
and has a determinant of
![\det\mathbf J=-\dfrac{2x}y](https://tex.z-dn.net/?f=%5Cdet%5Cmathbf%20J%3D-%5Cdfrac%7B2x%7Dy)
Note that we need to use the Jacobian in the other direction; that is, we've computed
![\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}](https://tex.z-dn.net/?f=%5Cmathbf%20J%3D%5Cdfrac%7B%5Cpartial%28u%2Cv%29%7D%7B%5Cpartial%28x%2Cy%29%7D)
but we need the Jacobian determinant for the reverse transformation (from
![(x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29)
to
![(u,v)](https://tex.z-dn.net/?f=%28u%2Cv%29)
. To do this, notice that
![\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28x%2Cy%29%7D%7B%5Cpartial%28u%2Cv%29%7D%3D%5Cdfrac1%7B%5Cdfrac%7B%5Cpartial%28u%2Cv%29%7D%7B%5Cpartial%28x%2Cy%29%7D%7D%3D%5Cdfrac1%7B%5Cmathbf%20J%7D)
we need to take the reciprocal of the Jacobian above.
The integral then changes to
![\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7B%5Cmathcal%20W_%7B%28x%2Cy%29%7D%7De%5E%7Bxy%7D%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D%5Ciint_%7B%5Cmathcal%20W_%7B%28u%2Cv%29%7D%7D%5Cdfrac%7Be%5Eu%7D%7B%7C%5Cdet%5Cmathbf%20J%7C%7D%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
Answer:
A. y = 10x +5
Step-by-step explanation:
For 0, is 5
For 1 is 15 so on