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disa [49]
3 years ago
15

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

Mathematics
1 answer:
KatRina [158]3 years ago
6 0

Answer: 8.944272

Step-by-step explanation:

Distance formula

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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:

<h3> (x-8)²+(y-3)² = 81</h3>

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
2 years ago
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