Answer:
C
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Here m = - 3, so
y = - 3x + c ← is the partial equation
T find c substitute (2, - 1) into the partial equation
- 1 = - 6 + c ⇒ c = - 1 + 6 = 5
y = - 3x + 5 ← in slope- intercept form
Add 3x to both sides
3x + y = 5 ← in standard form → C
Answer :Plotting the points into the coordinate plane gives us an observation that this quadrilateral with vertices d(0,0), i(5,5) n(8,4) g(7,1) is a KITE, as shown in figure a.
Step-by-step explanation:
Considering the quadrilateral with vertices
d(0,0)
i(5,5)
n(8,4)
g(7,1)
Plotting the points into the coordinate plane gives us an observation that this quadrilateral with vertices d(0,0), i(5,5) n(8,4) g(7,1) is a KITE, as shown in figure a.
From the figure a, it is clear that the quadrilateral has
Two pairs of sides
Each pair having two equal-length sides which are adjacent
The angles being equal where the two pairs meet
Diagonals as shown in dashed lines cross at right angles, and one of the diagonals does bisect the other - cuts equally in half
Please check the attached figure a.
The answer is c. When the added number is outside of the radical it adds to the y, moving it up
"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)
Yes there right the answer is three
