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abruzzese [7]
3 years ago
12

A school needs 2,400 notebooks for next year if the notebooks come in packages of 60, how many packages should the school buy?

Mathematics
2 answers:
coldgirl [10]3 years ago
6 0

Answer:

40

-step explanation:

Andrew [12]3 years ago
5 0

Answer:

40

Step-by-step explanation:

Divide 2400 by 60 to get 40

Hope this helps!

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What is the value of 128x3/7 as a whole number ?
UNO [17]

Answer:

<u><em>54.86 or 55, rounded to a whole number</em></u>

Step-by-step explanation:

(128*3) = 384

384/7 = 54.86

This is not a whole number

6 0
1 year 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
I need to know the steps on how to do it
Sergeu [11.5K]
You need help with what problem
4 0
3 years ago
I need some help please
igor_vitrenko [27]
The answer I got to this was 2.3 but I don't know if that's correct. 
8 0
3 years ago
Which equation is true for the value b=2 ?
katrin2010 [14]
The correct answer is B
8 0
2 years ago
Read 2 more answers
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