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Katyanochek1 [597]
2 years ago
10

The center of a circle and a point on the circle are given. Write the equation of the circle in standard form.

Mathematics
1 answer:
pogonyaev2 years ago
6 0

Answer:

A.Because ,center is (3,2)and point is (-2-3) .

and answer is (x-3)2+(y+2)2=52.

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Someone pls help me <br><br><br> Solve for x
mezya [45]
The correct answer is X=-2.

First remove the parentheses and factor out 2.

1/2•(2-6x)-4(x+3/2)=-(x-3)+4 turns into
1/2•2(1-3x)-4(x+3/2)=-(x-3)+4.
Then distribute -4 through the parentheses,
1/2•2(1-3x)-4x-6=-(x-3)+4
Then change the sign of the x-3 to x+3,
1/2•2(1-3x)-4x-6=-x+3+4

Then reduce the numbers with the greatest common factor which is 2,
and remove unnecessary parentheses,
1-3x-4x-6=-x+7
Then calculate the differences and collect like terms,
-5-7x=-x+7

Then move the variable to the left side and change its sign, and move the constant to the right hand side and change its sign,
-7x+x=7+5
Then add and collect the like terms,
-6x=12
And finally divide both sides by -6 to get X=-2.

Whew lol


8 0
2 years ago
Read 2 more answers
Number line solutions to n&gt;-2
docker41 [41]

Answer:

start at -2 then go right

Step-by-step explanation:

then leave a arrow mark me brainiest.

5 0
2 years ago
Help? im not really sure how to do this :(
ICE Princess25 [194]

Answer: doritos

Step-by-step explanation: mmmm doritos

4 0
3 years ago
Let φ(x, y) = arctan (y/x) .
Alexandra [31]

Answer:

a) \large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})

b) \large \mathbb{R}^2-\{(0,0)\}

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})

On one hand we have,

\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}

On the other hand,

\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}

So

\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})

b)

The domain of definition of F is  

\large \mathbb{R}^2-\{(0,0)\}

i.e., all the plane X-Y except the (0,0)

c)

Here we want to find all the points such that

\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)

where k is a real number other than 0.

But this means

\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

8 0
2 years ago
Is the system of equations is consistent, consistent and coincident, or inconsistent?
kicyunya [14]
<span><span>Hiya!

Y=−3x+1
</span><span>
2y=−6x+2

The answer is...

</span></span><span>consistent and coincident
</span>
Hope This Helps!
(If it Helps I took the test, its 100% Right)
(Brainliest is always appreciated)
If You Have Any More Questions Feel Free To Ask! :)
5 0
3 years ago
Read 2 more answers
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