Please answer the question 3x < 18

Mathematics

1 answer:

Ann [662]3 years ago

4 0

3x<18 I'm not sure what the rule for this problem is.

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Solve the inequalities -150x ≥ − 2,400 and -336 ≥ − 21y.

astra-53 [7]

Ok so easy peasy
remember that when divide or multiply by negative, reverse the inequality symbol example
2>3 times -1=
-2<-3 so

-150x>-2400
divide both sides by -150
flip sign
x<16

-336>-21y
divide both sides by -21
flip sign
16<y

they seem to have the same solution

x=y>16

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3 years ago

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Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

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3 years ago

4x^2+12xy+9y^2 tại x=2, y=2

Katena32 [7]

Step-by-step explanation:

hope it's helpful for you

PLS REFER THE ABOVE ATTACHMENT

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2 years ago