All categories

3 years ago

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

You might be interested in

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 $(2,10,\pm2\sqrt5)$ .