What isn’t he solution set of the quadratic inequality 4(x + 2)squared less than or equal to 0

3 answers
Question:

What isn’t he solution set of the quadratic inequality 4(x + 2)squared less than or equal to 0

Answers

a perfect square less than or equal to 16 is 16

Step-by-step explanation:

the perfect squares between 0 and 16 are

1 , 4 , 9 , 16

anyone of them can be your answer.

Looks like we're given

[tex]\vec F(x,y)=\langle-x,-y\rangle[/tex]

which in three dimensions could be expressed as

[tex]\vec F(x,y)=\langle-x,-y,0\rangle[/tex]

and this has curl

[tex]\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle[/tex]

which confirms the two-dimensional curl is 0.

It also looks like the region [tex]R[/tex] is the disk [tex]x^2+y^2\le5[/tex]. Green's theorem says the integral of [tex]\vec F[/tex] along the boundary of [tex]R[/tex] is equal to the integral of the two-dimensional curl of [tex]\vec F[/tex] over the interior of [tex]R[/tex]:

[tex]\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA[/tex]

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of [tex]R[/tex] by

[tex]\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle[/tex]

[tex]\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt[/tex]

with [tex]0\le t\le2\pi[/tex]. Then

[tex]\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt[/tex]

[tex]=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0[/tex]

You're looking for the extreme values of [tex]x^2+3y^2+13x[/tex] subject to the constraint [tex]x^2+y^2\le1[/tex].

The target function has partial derivatives (set equal to 0)

[tex]\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2[/tex]

[tex]\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0[/tex]

so there is only one critical point at [tex]\left(-\dfrac{13}2,0\right)[/tex]. But this point does not fall in the region [tex]x^2+y^2\le1[/tex]. There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of [tex]x^2+y^2\le1[/tex] by

[tex]x=\cos u[/tex]

[tex]y=\sin u[/tex]

with [tex]0\le u<2\pi[/tex]. Then [tex]t(x,y)[/tex] can be considered a function of [tex]u[/tex] alone:

[tex]t(x,y)=t(\cos u,\sin u)=T(u)[/tex]

[tex]T(u)=\cos^2u+3\sin^2u+13\cos u[/tex]

[tex]T(u)=3+13\cos u-2\cos^2u[/tex]

[tex]T(u)[/tex] has critical points where [tex]T'(u)=0[/tex]:

[tex]T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0[/tex]

[tex](1)\quad\sin u=0\implies u=0,u=\pi[/tex]

[tex](2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4[/tex]

but [tex]|\cos u|\le1[/tex] for all [tex]u[/tex], so this case yields nothing important.

At these critical points, we have temperatures of

[tex]T(0)=14[/tex]

[tex]T(\pi)=-12[/tex]

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

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