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borishaifa [10]
3 years ago
10

What the center of X^2+(y+3)=9

Mathematics
1 answer:
olga_2 [115]3 years ago
4 0
Assuming you mean x^2+(y+3)^2=9, which is the equation of a circle :

the general equation of a circle of radius R and center (a,b) is (x-a)^2+(y-b)^2=R^2 hence here the center is \boxed{(0,-3)}
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Which square root would be classified as rational √2, √16. √24, √40 which one of these numbers' square root be rational
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Square root of 16 would be the only rational number since it is the only whole number.  
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3 years ago
Like charges repel and unlike charges attract. Coulomb’s law states that the force F of attraction or repulsion between two char
Artist 52 [7]

Step-by-step explanation:

The electrical force between two unlike charge is given by :

F=\dfrac{kq_1q_2}{r^2}

k is electrostatic constant

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It is clear that the electric force is inversely proportional to the square of the distance between them.

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8 0
3 years ago
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Fruit juice in 4/5 pint cartons is sold by the park in 5/6 pint mugs. How many mugs are in a​ carton?
Alinara [238K]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit. (Hint: Assume that
Vadim26 [7]
So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
that simply means, 3 circles with a radius of 3, overlapping the triangle

now, the area in the middle, the shaded one, will be, the whole area of the triangle MINUS those 3 circle sectors

hmmmm each sector has 60°, that means, all three of them will then be 60+60+60 or 180°, so the area of those three sectors, can be combined into a 180° sector, well, hell, 180° is really half a circle

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so    \bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
\textit{area of half a circle}\\\\
A=\cfrac{\pi r^2}{2}\\\\
-----------------------------\\\\

\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
\end{array}\impliedby \textit{area of shaded area}\\\\
-----------------------------\\\\
\boxed{\cfrac{36\sqrt{3}}{4}-\cfrac{9\pi }{2}}

you can, add the fractions if you want, or leave them like that, or get their difference by using their decimal format

8 0
3 years ago
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13 is the answer.
7 0
4 years ago
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