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3 years ago

If x = -6,which inequiality is true

Mathematics

1 answer:

zimovet [89]3 years ago

7 0

The only equation that works with x=-6 is A

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Plz someone answer this with an explanation

worty [1.4K]

Answer:

1/3

Step-by-step explanation:

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3 years ago

Find the gradient of the line segment between points (8,6) and (10,14)

fenix001 [56]

Answer: 4

Step-by-step explanation:

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2 years ago

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

3 years ago

Find the interval(s) of upward concavity on this accumulation function.

maxonik [38]

$\displaystylef(x)=\int_{0}^{x^2}\sec^2(\sqrt{x})dx \\=\int_{0}^{\sqrt{x^2}}\sec^2(u)\cdot2udu \\=2\int_{0}^{\sqrt{x^2}}\sec^2(u)du \\=2\Big[u\tan(u)-\int\tan(u)du\Big]_{0}^{\sqrt{x^2}} \\=2\Big[u\tan(u)+\ln\Big(\mathrm{abs}(\cos(u))\Big)\Big]_0^x \\=2\Big(x\tan(x)+\dfrac{1}{2}\ln\Big(\cos^2(x)\Big)\Big) \\=2x\tan(x)+\ln(\cos^2(x))$

Hope this helps.

6 0

3 years ago

Ricard lost 6 golf balls while playing yesterday. He bought a box of 12 golf balls ,then lost 4 on the course today . He now has

viktelen [127]

16, first you need to add on the four he lost to the 18 he ended with.

4 0

3 years ago