Answer:
a. -2
b. s = 2 and y = -5
c. (2.5, -9)
d. -7x - 2
Step-by-step explanation:
a. In the image, I have used rise over run, which is -8/4 or -2.
b. slope formula is mx + b = y, where m is slope and b is the y-intercept. In the equation 2x - 5y = -10, 2 is the slope and -5 is the y-intercept.
c. Plug in all given numbers into an equation and the ones you do not have will be in the points you need it to be. Once you graph the equation you will receive (2.5, -9).
d. y = mx + b
5 = (-7)(-1) + b
5 = 7 + b
-2 = b
y = -7x - 2
Answer:
The length of AA' = √29 = 5.39
Step-by-step explanation:
* Lets revise how to find the length of a line joining between
any two points in the coordinates system
- If point A is (x1 , y1) and point B is (x2 , y2)
- The length of AB segment √[(x2 - x1)² + (y2 - y1)²]
* Lets use this rule to solve the problem
∵ Point A is (0 , 0)
∵ Point A' = (5 , 2)
∵ (x2 - x1)² = (5 - 0)² = 5² = 25
∵ (y2 - y1)² = (2 - 0)² = 2² = 4
∴ The length of AA' = √(25 + 4) = √29 = 5.39
The scale factor is what u multiply the first image by to get the second image. So by dividing the second image by the first image will give u the scale factor.
6 / 1.5 = 4....so there is a scale factor of 4 because 1.5 * 4 = 6
Answer: 8.1 $
Step-by-step explanation: 18 percent of 45 is 8.1 ez math hope this helps.
![\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.
