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3 years ago

Find the equation of the circle in standard form for the given center (h,k) and radius r: (h,k)=(0,0), r=9

Mathematics

2 answers:

Vsevolod [243]3 years ago

8 0

X^2 + y^2 = 81 is the equation of a circle you specified

professor190 [17]3 years ago

6 0

Answer: $x^2+y^2=81$

Step-by-step explanation:

The equation of circle ( in general form ) is given by :_

$(x-h)^2+(y-k)^2=r^2$

, where r= radius of the circle .

(h,k) = center of the circle .

Given : Center = (h,k)= (0,0)

Radius = r = 9 units

Then the equation of the circle will be :-

$(x-0)^2+(y-0)^2=(9)^2$

$\Rightarrow x^2+y^2=81$

Hence, the required equation of the circle = $x^2+y^2=81$

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The mistake Jadyn made in the long division is

Mistake in subtraction
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1 year ago

Find the absolute maximum and minimum values of f on the set D. f(x, y) = x3 − 3x − y3 + 12y + 1, D is quadrilateral whose verti

mojhsa [17]

$D$ is the set of points,

$\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}$

Check for critical points:

$f(x,y)=x^3-3x-y^3+12y+1\implies\begin{cases}f_x=3x^2-3=0\implies x=\pm1\\f_y=-3y^2+12=0\implies y=\pm2\end{cases}$

Of these 4 points, only 2 belong to $D$ , (-1, 2) and (1, 2), for which we have

$\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}$

Now look for extrema along the boundary.

If $x=-2$ , then

$f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2$

We have $f'(-2,y)>0$ for $-2$ and $f'(-2,y)$ for $2$ , which indicates a local maximum at $y=2$ and minima at the endpoints of this boundary. So

$\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}$

If $x=2$ , then

$f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2$

We have $f'(2,y)$ for $2$ , so we have extrema at the endpoints of this boundary.

$\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}$

If $y=x$ , then

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which tells us $f$ is strictly increasing on this boundary, giving the extrema we already know about,

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If $y=3$ , then

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We have $f'(x,3)>0$ for $-2$ and $1$ , and $f'(x,3)$ for $-1$ . This indicates a maximum at $x=-1$ and a minimum at $x=1$ , with

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From this analysis, we find that $f$ attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).

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