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irina1246 [14]
3 years ago
5

A cell tower is located at and transmits a circular signal that covers three major cities. The three cities are located on the c

ircle and have the following coordinates: , , and . Find the equation of the circle.
Mathematics
1 answer:
Lelu [443]3 years ago
4 0

Answer:

(x+8)^2 +(y-4)^2 = 25

Step-by-step explanation:

Assuming this complete problem: "A cell tower is located at (-8, 4) and transmits a circular signal that covers three major cities. The three cities are located on the circle and have the following coordinates: G (-4, 7), H (-13, 4), and I (-8, -1). Find the equation of the circle"

For this case the generla equation for the circle is given by:

(x-h)^2 +(y-k)^2 = r^2

From the info we know that the tower is located at (-8, 4) so then h = -8 and k = 4, so then we need to find the radius. So we have the equation like this:(x+8)^2 +(y-4)^2 = r^2

If the 3 points are on the circle then satisfy the equation. We can use the first point (-4,7) and if we replace we can find the value for r^2

(-4+8)^2 +(7-4)^2 =25= r^2

So then r = \sqrt{25}=5

And if we replace the second point we got this:

(-13+8)^2 +(4-4)^2 = 25 =r^2

And for the third point we have:

(-8+8)^2 +(-1-4)^2 = 25 =r^2

And we got the same result.

So then our final equation is given:

(x+8)^2 +(y-4)^2 = 25

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Answer:

27/77

Step-by-step explanation:

I hope this is correct

Have a good Day!

3 0
3 years ago
Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
ycow [4]

This  involves quite a lot of arithmetic to do manually.

The first thing you do is to make the first number in  row 2  = to 0.

This is done by R2 = -3/2 R1 + R2

so the matrix becomes

( 2        1          1)    ( -3 )

( 0    -13/2   3/2)   (1/2 )

(5       -1           2)  (-2)

Next step is to make  the 5 in row 5  = 0  

then  the -1  must become zero

You aim  for the form

( 1 0 0) (x)

(0 1 0) (y)

(0 0 1) ( z)

x , y and z will be the required solutions.


4 0
3 years ago
Read 2 more answers
) find a vector parallel to the line of intersection of the planes 5x − y − 6z = 0 and x + y + z = 1.
snow_tiger [21]
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.

Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>

We calculate the cross product as a determinant of (i,j,k) and the normal products

    i   j   k
   5 -1 -6
   1  1  1

=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>

Check orthogonality with normal vectors using scalar products
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<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0

Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
5 0
3 years ago
Solve using long division.
yuradex [85]
I pray to hod for yourgood day
5 0
3 years ago
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A circle has an area of 40 square inches. Find the radius to 2 decimal places.
Oksanka [162]

The area of a circle is given by

A=\pi r^2

We can solve this equation for r: divide both sides by pi:

r^2=\dfrac{A}{\pi}

Consider the (positive) square root of both sides:

r = \sqrt{\dfrac{A}{\pi}}

So, in your case, we have

r = \sqrt{\dfrac{40}{\pi}}\approx 3.57

7 0
3 years ago
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