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lubasha [3.4K]
3 years ago
9

Find the mean. 217, 230, 214, 227, 196, 235, 220, 224, 208, 209, 191, 205, 184, 214, 219, 208, 227, 194, 228, 186, 201, 239

Mathematics
1 answer:
xeze [42]3 years ago
6 0

Answer:

212.54

Step-by-step explanation:

To find the mean, we take the sum of all numbers and divide by the total amount in the set

Total sum: 4676

Numbers in the set: 22

4676 divided by 22 = approx. 212.54

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alekssr [168]

A = 1/2 bh

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The height of a triangle is equal to twice the area A divided by the base ,b.

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vesna_86 [32]
Perimeter = 2w + 2l
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Use distributive property
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raketka [301]
Answer is:
x ≥ -18/5

2+4/9x ≥ 4+x
4/9x-x ≥4-2
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- 5/9x ≥2 multiple by 9
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6 0
3 years ago
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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
I NEED HELP SOMEONE PLEASE
slavikrds [6]

Answer: 45


Step-by-step explanation: 21 x 3 = 63, because 3x3 is 9. 9 tickets with a $2 discount would mean you take off $18, so I believe the answer would be 63-18=45.


5 0
3 years ago
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