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svlad2 [7]
3 years ago
15

Compare the first and last verses of the song "Footprints."

Mathematics
1 answer:
musickatia [10]3 years ago
4 0

Answer:

It helps express the song's central idea.

Step-by-step explanation:

I took the quiz

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Create a data set that contains 8 to 12 values such that the mean is greater than the median
rewona [7]

8, 8, 8, 8, 8, 9, 9, 9, 9 would work as a sequence.


Median: 8

Mean: 8.4444....

5 0
3 years ago
What is the equation for the points (-3,1) and (9,7)
Scorpion4ik [409]

Answer:

The equation of line with given points is 2Y - X - 5 = 0

Step-by-step explanation:

Given points are ( - 3 , 1)   and   (9 , 7)

Equation of line is y = mx +c

where m is the slop of line

Now m = \frac{y2 - y1}{x2 - x1}

Or,    m = \frac{7 - 1}{9 + 3}

so, slop = \frac{6}{12}

∴   slop = \frac{1}{2}

Now the equation of line with points ( -3 , 1) and slop m is :

Y - y1 = m ( X - x1)

Or, Y - 1 =  \frac{1}{2} (X + 3)

Or, 2Y - X - 5 = 0

Hence The equation of line with given points is 2Y - X - 5 = 0 Answer

6 0
3 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
Please help! <br> (sorry about the cracks in my ipad)
Talja [164]

Answer:

4

Step-by-step explanation:

6 0
3 years ago
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3x(2ab - b) - 2y(2a - b)<br>please i need help!!​
Salsk061 [2.6K]

Answer:

6abx-3bx-4ay+2by

Step-by-step explanation:

3 0
3 years ago
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