Answer:
r = √13
Step-by-step explanation:
Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms: x^2 + 6x + y^2 -2y = 3. Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."
We need to "complete the square" of x^2 + 6x. I'll just jump in and do it: Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x: x^2 + 6x + 3^2 - 3^2. Then do the same for y^2 - 2y: y^2 - 2y + 1^2 - 1^2.
Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2. Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1. Making these replacements:
(x + 3)^2 - 9 + (y - 1)^2 -1 = 3. Move the constants -9 and -1 to the other side of the equation: (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13
Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r: r^2 = 13, so that the radius is r = √13.
The 5 and 10 min cross. 10
<span>The 5 min returns. 15 </span>
<span>The 20 and 25 min cross 40 </span>
<span>The 10 min comes back 50 </span>
<span>The 5 and 10 min cross again in60 Mins</span>
Answer:
(2,-3)
Step-by-step explanation:
I am not sure if you meant the first equation to be y or -y. I solved it as y.
y = x-5 -x -3y =7
I am going to take the second equation and write it as x =
-x - 3y = 7 Give equation
-x = 3y +7 Add 3y to both sides
x = -3y-7 Multiplied each term in the equation by -1 so that x could be positive
I am going to substitute -3y-7 for x in the first equation up above
y = x - 5
y = -3y -7 - 5 Substitute -3y-7 for x
y = -3y -12 Combined -7-5
4y = -12 Added 3y to both sides
y = -3 Divided both sides by 4.
I now know that y is -3, I will plug that into x = -3y-7 to solve for x
x = -3(-3) -7
x = 9-7 A negative times a negative is a positive
x = 2
You can do 12/12 8/8 10/10 3/3 and 2/2 all equal to 6/6 all in different ways.
This the concept of geometry, we are required to get the coordinates of the point p which is the center of the circle.
The center of the circle will be at the point:
(x,y)
=[(6-1)/2,(6-3)/2]
=(5/2,3/2)
