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1 year ago

I can send the other options? I just need this explained?

Mathematics

1 answer:

OLEGan [10]1 year ago

3 0

The standard equation of a circle is given by:

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

where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.

For:

$\begin{gathered} (x+3)^2+(y+1)^2=16 \\ \end{gathered}$

Rewrite the equation as:

$\begin{gathered} (x-(-3))^2+(y-(-1))^2=4^2 \\ so\colon \\ h=-3 \\ k=-1 \\ r=4 \end{gathered}$

We can conclude it is a circle with center at (-3,-1) and radius of 4

You might be interested in

14 + 2y - 3+ 7x - 12y +3x - 20=

Trava [24]

Answer:

The answer is 10x - 10y - 9

Step-by-step explanation:

14 - 3 - 20 = -9

2y - 12y = -10y

7x + 3x = 10x

14 + 2y - 3 + 7x - 12y + 3x - 20 = 10x - 10y - 9

4 0

3 years ago

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

vichka [17]

Answer:

A(-1,0) is a local maximum point.

B(-1,0) is a saddle point

C(3,0) is a saddle point

D(3,2) is a local minimum point.

Step-by-step explanation:

The given function is

$f(x,y)=x^3+y^3-3x^2-3y^2-9x$

The first partial derivative with respect to x is

$f_x=3x^2-6x-9$

The first partial derivative with respect to y is

$f_y=3y^2-6y$

We now set each equation to zero to obtain the system of equations;

$3x^2-6x-9=0$

$3y^2-6y=0$

Solving the two equations simultaneously, gives;

$x=-1,x=3$ and $y=0,y=2$

The critical points are

A(-1,0), B(-1,2),C(3,0),and D(3,2).

Now, we need to calculate the discriminant,

$D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$

But, we would have to calculate the second partial derivatives first.

$f_{xx}=6x-6$

$f_{yy}=6y-6$

$f_{xy}=0$

$\Rightarrow D=(6x-6)(6y-6)-0^2$

$\Rightarrow D=(6x-6)(6y-6)$

At A(-1,0),

$D=(6(-1)-6)(6(0)-6)=72\:>\:0$ and $f_{xx}=6(-1)-6=-18\:$

Hence A(-1,0) is a local maximum point.

See graph

At B(-1,2);

$D=(6(-1)-6)(6(2)-6)=-72\:$

Hence, B(-1,0) is neither a local maximum or a local minimum point.

This is a saddle point.

At C(3,0)

$D=(6(3)-6)(6(0)-6)=-72\:$

Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.

At D(3,2),

$D=(6(3)-6)(6(2)-6)=72\:>\:0$ and $f_{xx}=6(3)-6=12\:>\:0$

Hence D(3,2) is a local minimum point.

See graph in attachment.

3 0

3 years ago

2.) Two students are working on a simplification problem. Their work is shown below. Is either of

kvasek [131]

Answer:

Both Jeff and Lori are wrong.

Step-by-step explanation:

$3 - \frac{(4 \times 3)^{2} }{2} + (1 \times 6)$

$3 - \frac{12^{2}}{2} + 6$

$3 - \frac{144}{2} + 6$

$3 - 72 + 6$

$- 72 + 9$

$- 63$

-63 is the final answer

8 0

3 years ago

Fill in the blanks to rewrite the following statement with variables: Given any positive real number, there is a positive real n

sveticcg [70]

Answer:

a) there is s such that r>s and s is positive

b) For any r>0 , there exists s>0 such that s<r

Step-by-step explanation:

a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.

b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.

3 0

3 years ago

Can you find the midpoint of AB

Masteriza [31]

Use the distance formula for the 2 points and then halve the answer

7 0

2 years ago

Read 2 more answers