"determine the location" or namely, is it inside the circle, outside the circle, or right ON the circle?
well, we know the center is at (1,-5) and it has a radius of 5, so the distance from the center to any point on the circle will just be 5, now if (4,-1) is less than that away, is inside, if more than that is outiside and if it's exactly 5 is right ON the circle.
well, we can check by simply getting the distance from the center to the point (4,-1).
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[4-1]^2+[-1-(-5)]^2}\implies d=\sqrt{(4-1)^2+(-1+5)^2} \\\\\\ d = \sqrt{3^2+4^2}\implies d =\sqrt{9+16}\implies d=\sqrt{25}\implies \stackrel{\textit{right on the circle}}{d = 5}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%5Cstackrel%7Bcenter%7D%7B%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-5%7D%29%7D%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B%5B4-1%5D%5E2%2B%5B-1-%28-5%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%284-1%29%5E2%2B%28-1%2B5%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B3%5E2%2B4%5E2%7D%5Cimplies%20d%20%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20d%3D%5Csqrt%7B25%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bright%20on%20the%20circle%7D%7D%7Bd%20%3D%205%7D)
There are 5. In 5000, there are 5 thousandths.
~Silver
Answer:
y= (1/2)x+(3/4)
Step-by-step explanation:
Answer:
y = -2(x - .5)^2 - 2.5
Step-by-step explanation:
You first want to find the vertex of the parabola, which is found at (.5, -2.5)
You then want to add this to the function "y = x"
y = (x - .5) - 2.5
You would then notice this would need to be a quadratic formula, so you would add the squared to the equation, and make the leading coefficient negative because it is facing downwards.
y = -(x - .5)^2 - 2.5
Finally, you would notice that the graph has a y-intercept of -3. You can Vertically stretch the equation y a factor of 2 to finish the graph.
y = -2(x - .5)^2 - 2.5
4/12 is equal to 0.3 in decimal form.
