"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)
Answer:
Step-by-step explanation:
f(x) = 2x + 5
f(-6)= 2(-6) + 5 = -12 + 5 = -7
Answer:
Slope
Step-by-step explanation:
The point of slope is that it is the same the entire length of the line.
Answer:
It is a perfect square. Explanation below.
Explanation:
Perfect squares are of the form
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
. In polynomials of x, the a-term is always x.(
(
x
+
c
)
2
=
x
2
+
2
c
x
+
c
2
)
x
2
+
8
x
+
16
is the given trinomial. Notice that the first term and the constant are both perfect squares:
x
2
is the square of x and 16 is the square of 4.
So we find that the first and last terms correspond to our expansion. Now we must check if the middle term,
8
x
is of the form
2
c
x
.
The middle term is twice the constant times x, so it is
2
×
4
×
x
=
8
x
.
Okay, we found out that the trinomial is of the form
(
x
+
c
)
2
, where
x
=
x
and
c
=
4
.
Let us rewrite it as
x
2
+
8
x
+
16
=
(
x
+
4
)
2
. Now we can say it is a perfect square, as it is the square of
(
x
+
4
)
.
