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Ksivusya [100]
3 years ago
8

Find square root of 7921​

Mathematics
2 answers:
V125BC [204]3 years ago
8 0

Answer: 89

89 times 89 is 7921. Since 7921 is a 4 digit perfect square, it has 89 as its root. Square root of 7921=89

Bas_tet [7]3 years ago
6 0

Answer:89

Step-by-step explanation:

√7921

89|7921

89|89

√89×89=89

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Michael has 13 1/4 pizzas left in the freezer at his pizza shop. The next day a group comes in and orders 5 1/3 pizzas how many
Papessa [141]

Answer:

8 1/12

Step-by-step explanation:

13 1/4 - 5 1/3  = (13-5) + (1/4 - 1/3)

                      = 8 + (4/12 - 3/12)

                      = 8 1/12

8 0
4 years ago
I just need to find this data set out, please anyone?​
rewona [7]

Answer:

Ok so...

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Step-by-step explanation:

3 0
3 years ago
Someone please help me with this !!!
a_sh-v [17]

Answer:

what does it say young buck

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
4 years ago
I hate math bro please help me ill mark u brainliest
yulyashka [42]
Please mark me brainless :)
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3 years ago
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