Hi please help this is due today thank u so much.

It says to place points on them but i still don't get it. help please.

xxTIMURxx [149]

Answer: A'=(1, 3); B'=(-3, 4);C'=(3, 0); D'=(-2, 5)

You can check the PNG attached as well.

Step-by-step explanation:

You need to represent the symmetry of every given points respet to the line

$y = 2$

In that case, the line beeing paralell to the x- axis, x- value of the symmetry is the same of the given point and y = 2 is the middle between both points.

Point A(1, 1)

$x_{A} = 1\\ x_{A'} = 1 \\\\\frac{y_{A} +y_{A'} }{2} =2\\y_{A'} = 4 - y_{A} = 4 - 1 = 3$

Point B(-3, 0)

$x_{B} = 1\\ x_{B'} = 1 \\\\\frac{y_{B} +y_{B'} }{2} =2\\y_{B'} = 4 - y_{B} = 4 - 0 = 4$

Point C(3, 4)

$x_{C} = 1\\ x_{C'} = 1 \\\\\frac{y_{C} +y_{C'} }{2} =2\\y_{C'} = 4 - y_{C} = 4 - 4 = 0$

Point D(-2, -1)

$x_{D} = 1\\ x_{D'} = 1 \\\\\frac{y_{D} +y_{D'} }{2} =2\\y_{D'} = 4 - y_{D} = 4 - (-1) = 4 + 1 = 5$

4 0

3 years ago

A linear function contains the following point

Vsevolod [243]

Answer:

D

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What equation describes the circle that is centered at (8, 3) and has a radius of 9

Sloan [31]

Answer:

Step-by-step explanation:

The general equation of a circle is expressed as;

(x-a)²+(y-b)² = r² where;

(a, b) is the centre of the circle

r is the radius

Given the centre (8, 3) and the radius of 9, the equation of the circle will be;

(x-a)²+(y-b)² = r²

(x-8)²+(y-3)² = 9²

(x-8)²+(y-3)² = 81

hence the required equation is (x-8)²+(y-3)² = 81

5 0

2 years ago

Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

5 0

3 years ago

Complete the statement using always, sometimes, or never.

babunello [35]

it had to be b cuz now all quadrilateral have straight sides

3 0

3 years ago