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ki77a [65]
2 years ago
8

Fill in the blank.

Mathematics
1 answer:
Semenov [28]2 years ago
6 0

O centroid

Step-by-step explanation:

In geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. Every triangle has three medians, one from each vertex, and they all intersect each other at the triangle's centroid.

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A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.
olga55 [171]
We know that:

(x-a)^2+(y-b)^2=r^2

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}

and all we have to do is solve it for a, b and r.

There will be:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\


From equations (II) and (III) we have:

\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}

and from (I) and (II):

\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}

Now we can easly calculate a and b:

\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}

Finally we calculate r^2:

a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}

And the equation of the circle is:

(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}
7 0
3 years ago
please let us to do with it and it was the only one that I can get the best way for you and I am a bit more than one person who
Lelechka [254]
Inside, Lighthouse, Overboard (If You Can go Diagonal) Outback, Outdoors
Hope that helps :)
8 0
3 years ago
PLS HELP WILL MARK BRAINLIEST IM SO CLOSE TO FINISHING
Tom [10]
The answer is the 3 one
3 0
3 years ago
Read 2 more answers
Can someone explain interval notations? Also, How do you figure them out?
sergeinik [125]

A notation for representing an interval as a pair of numbers. The numbers are the endpoints of the interval. Parentheses and/or brackets are used to show whether the endpoints are excluded or included.

For example, [3, 8) is the interval of real numbers between 3 and 8, including 3 and excluding 8.


8 0
2 years ago
HELP PLEASEEE. I DONT UNDERSTAND
Shtirlitz [24]

Answer:

D

Step-by-step explanation:

So we have the equation:

x^2-6x+58=0

And we want to solve for x.

We can solve it by completing the square.

First, subtract 58 from both sides:

x^2-6x=-58

Divide the b term by 2 and square it:

(-6)/2=-3\\(-3)^2=9

So, add 9 to both sides:

(x^2-6x+9)=-58+9

On the left, the perfect square trinomial pattern. Add on the right. So:

(x-3)^2=-49

Take the square root of both sides:

x-3=\pm \sqrt{-49}

The square root of -49 is 7i:

x-3=\pm 7i

Add 3 to both sides:

x=3\pm 7i

So, our answer is D.

And we're done!

4 0
2 years ago
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