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larisa86 [58]
3 years ago
15

Please answer! Very easy and will mark brainlist.

Mathematics
2 answers:
kondor19780726 [428]3 years ago
5 0

Answer:

18.57

Step-by-step explanation:

First you have to simplify the equation using cross-multiplication

You should get 7v=130

Then divide both sides by 7

You should get v=130/7

simplify that and get 18 4/7

Turn that into decimal form and you get 18.57

AleksAgata [21]3 years ago
4 0

Answer:

18.57

Step-by-step explanation:

1. Multiply all terms by the same value to eliminate fraction denominators

v/10 = 13/7

10 x v/10 = 10 x 13/7

2. Simplify

130/7 = 18.571428571429

3. Round to the Nearest Tenth

18.57

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The further away ( to the left) from 0 a negative number is the smaller it is.

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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
What property is this: 3 * 1/3 = 1
Rama09 [41]

Answer:

Inverse property of multiplication

3 0
2 years ago
-56 ÷ c = 7, c =<br> pls help me
Rus_ich [418]

Answer:

-392

Step-by-step explanation:

-56 x 7=-392

-392 / -56=7

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3 years ago
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