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inna [77]
3 years ago
8

A group of point that lie on the same plane are called

Mathematics
1 answer:
melamori03 [73]3 years ago
5 0

Hey There!

Answer Coplanar Points

The term to describe a group of points that lie on the same plane are called.

Coplanar Points

If the points were to lie on the same line, the term to describe this would be called.

Collinear point

Hope this helped!

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Which of the following statements are true of literal equations?
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Answer:

1. They consist primarily of variables.

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4. They often describe real-world relationships.

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3 years ago
BRAINLIEST ANSWERS ONLY.
mars1129 [50]

Answer:

Heyy

Step-by-step explanation:

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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).
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3 years ago
Answer these problems please! Solving for x!
Tema [17]

Answer:

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Step-by-step explanation:

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2 years ago
Please help, it would be much appreciated.A certain state uses the following progressive
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Answer:

$3,300

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5.5% → 0.055

60,000 * 0.055 = $3,300

6 0
2 years ago
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