<p>I think it is the 2nd diagnostic test's 2nd and 6th question I don't get</p>
<p>so if you have the book</p>
<p>the first one's answer is circle, I don't get it visually</p>
<p>also, the 6th one's answer is 49</p>
<p>how did they get (y-5)...they redid the circle equation, how?</p>
<p>there's only one diagnostic...???</p>
<p>what page is it on?</p>
<p>test one in the back:)</p>
<p>I was doing that practice test today and was stumped as it had been quite some time since Alg 2
so
x^2+y^2-10y-36=0
so move 36 to the right
x^2+y^2-10y=36
no you have a -10y you need to get rid of and make a (y-sth)^2
So complete the square
divide -10 by 2 and square the term. You get 25. and 25 to both sides.
you get x^2+y^2-10y+25=61
so now factor and you get
x^2+(y-5)^2=61
so the radius in the square root of 61.
and circumference is 2piR so 2pi*square root of 61 and you get 49.
Hope that helped.</p>
<p>I'm assuming you meant question 3, since the answer is a circle.</p>
<p>first, draw the equilateral triangle, and choose two points. the set of points that are equidistant from those 2 points would form a line the perpendicularly bisects the triangle. now you're working on a 2D surface, so remeber the and points coming towards you, or going away from you on that line are also equidistant. So with the first condition you have a Plane that perpendicularly bisects the line that connect those 2 initially selected points.</p>
<p>Now with the 3rd point, the set is all points 2 inches from the vertex. When you draw it, you'd probablt draw a circle and answer c is the tempting answer, but you have to remember that they are taking a 3D space into account. so this circle is actually a sphere.</p>
<p>now the plane intersects all the points of a slice of the sphere. the plane through the sphere forms a circle.</p>
<p>-Aaron</p>