![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r}\\\\ -------------------------------\\\\ (x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~ \begin{cases} \stackrel{center}{(-1,0)}\\ \stackrel{radius}{6} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%7D%7B%20r%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%28x%2B1%29%5E2%2By%5E2%3D36%5Cimplies%20%5Bx-%28%5Cstackrel%7Bh%7D%7B-1%7D%29%5D%5E2%2B%5By-%5Cstackrel%7Bk%7D%7B0%7D%5D%5E2%3D%5Cstackrel%7Br%7D%7B6%5E2%7D~~~~%0A%5Cbegin%7Bcases%7D%0A%5Cstackrel%7Bcenter%7D%7B%28-1%2C0%29%7D%5C%5C%0A%5Cstackrel%7Bradius%7D%7B6%7D%0A%5Cend%7Bcases%7D)
so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2} \\\\\\ d=\sqrt{0+1}\implies d=1](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AA%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B-1-%28-1%29%5D%5E2%2B%281-0%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B1%29%5E2%2B1%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B0%2B1%7D%5Cimplies%20d%3D1)
well, the distance from the center to A is 1, namely is "inside the circle".
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2} \\\\\\ d=\sqrt{0+36}\implies d=6](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AB%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B-1-%28-1%29%5D%5E2%2B%286-0%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B1%29%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B0%2B36%7D%5Cimplies%20d%3D6)
notice, the distance to B is exactly 6, and you know what that means.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2} \\\\\\ d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AC%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B4-%28-1%29%5D%5E2%2B%5B-8-0%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%284%2B1%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B25%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B89%7D%5Cimplies%20d%5Capprox%209.43398)
notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.
Answer:
is this actually a question
Step-by-step explanation:
I suggest you look at the work in the pictures I attached. It examplains how you work out the problem and what the answer is
The area of a pyramid is the sum of area of the base and the area of slant triangles.
The area of the base is 1.5 × 1.5 = 2.25 cm²
The slant height of each triangle, s is 3 cm
The area of the triangles = 1/2 × 1.5 × 3 × 4
= 9 cm²
Therefore, the area of the pyramid
=9 +2.25
= 11.25 cm²
Answer:
y = -2/3x - 2
Step-by-step explanation:
Parallel lines have the same slope so slope = -2/3
Given (-3,0)
Slope-intercept: y = mx + b
0 = -2/3(-3) + b
0 = 2 + b
b = -2
then y = -2/3x - 2
