The distance between (2,0) and (5, -1) is

Mathematics

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

(3, -1)

Step-by-step explanation:

5-2=3

0-1=-1 (keep 0, change - to a +, flip 1 to a -1)

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Evaluate the spherical coordinate integral

expeople1 [14]

Rewrite the equations of the given boundary lines:

y = -x + 1 ==> x + y = 1

y = -x + 4 ==> x + y = 4

y = 2x + 2 ==> -2x + y = 2

y = 2x + 5 ==> -2x + y = 5

This tells us the parallelogram in the x-y plane corresponds to the rectangle in the u-v plane with 1 ≤ u ≤ 4 and 2 ≤ v ≤ 5.

Compute the Jacobian determinant for this change of coordinates:

$J=\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}1&1\\-2&1\end{bmatrix}\implies|\det J|=3$

Rewrite the integrand:

$-3x+4y=-3\cdot\dfrac{u-v}3+4\cdot\dfrac{2u+v}3=\dfrac{5u+7v}3$

The integral is then

$\displaystyle\iint_R(-3x+4y)\,\mathrm dx\,\mathrm dy=3\iint_{R'}\frac{5u+7v}3\,\mathrm du\,\mathrm dv=\int_2^5\int_1^45u+7v\,\mathrm du\,\mathrm dv=\boxed{333}$