We are given original equation: 
We need to find the enter and radius of a circle using the completing the square method.
The steps are as following :
Step 1 [original equation]: x^2 − 10x + y^2 + 12y = 20 .
Step 2 [group like terms]: (x^2 − 10x) + (y^2 + 12y) = 20
Step 3 [complete the quadratics]: (x^2 − 10x + 25) + (y^2 + 12y + 36) = 20 + (25 + 36).
Step 4 [simplify the equation]: (x^2 − 10x + 25) + (y^2 + 12y + 36) = 64.
Step 5 [factor each quadratic]: (x − 5)^2 + (y + 6)^2 = 8^2
Step 6 [identify the center and radius]: Center = (5, −6) Radius = 8.
<h3>Step 6 is incorrect.</h3><h3>The center should be (5,-6).</h3><h3> Replace − 5 with + 5 and replace + 6 with − 6.</h3>
Answer
Find out the length of OP .
To prove
As given
In △JKL, JO=44 in.
Now as shown in the diagram.
JP , MK, NL be the median of the △JKL and intresection of the JP , MK, NL be O .
Thus O be the centroid of the △JKL .
The centroid divides each median in a ratio of 2:1 .
Let us assume x be the scalar multiple of the OP and JO .
As given
JO = 44 in
2x = 44
x = 22 in
Thus the length of the OP IS 22 in .
Answer:
25.6 units
Step-by-step explanation:
From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).
First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:
where
are the coordinates of the first point
are the coordinates of the second point
- For AB:
![d=\sqrt{[1-(-5)]^{2}+(4-4)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5B1-%28-5%29%5D%5E%7B2%7D%2B%284-4%29%5E2%7D)
- For BC:
- For AC:
![d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5B3-%28-5%29%5D%5E%7B2%7D%20%2B%28-4-4%29%5E%7B2%7D%7D)
Next, now that we have our lengths, we can add them to find the perimeter of our triangle:
We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.
Answer:10
Step-by-step explanation: divide 60 by 6
Answer:
Step-by-step explanation: 58 - 3 = 55.
