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alexira [117]
3 years ago
13

What is 5x10 I'll give you brainly

Mathematics
2 answers:
levacccp [35]3 years ago
8 0

Answer:

50

Step-by-step explanation:

Because you just put the zero behind the 5 and it becomes 50

Yuki888 [10]3 years ago
4 0

Answer:

50

Step-by-step explanation:

Fifty lol Gosh why does my answer have to be 20 characters

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3) Of the 20 cookies at the bake sale, 35% are gingerbread. How many
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Answer:

the answer is 8. 75.

Step-by-step explanation:

25×35% = 8.75

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3 years ago
Help, please!!!<br><br> Simplify. (4/9)^2=
alina1380 [7]

Answer:

Exact form: \frac{16}{81}

Decimal form: 0.1975308

8 0
2 years ago
In the figure below an angle that is supplementary to angel FAB is:
Vlad1618 [11]

Answer:

CAB

Step-by-step explanation:

Angle CAB and Angle FAB together make 180 degrees(straight angle) . A supplementary angles equal to 180 degrees.

3 0
4 years ago
What would be the best way to get an unbiased sample that represents the population for the following topic: How do people feel
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2. random samples are the least biased samples.
4 0
4 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
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