that's what I got, you can confirm if it's correct or not... I'm not sure
have a nice dayʘ‿ʘ
Answer: look at the picture
Step-by-step explanation: Hope this help :D, Can I have a Brainliest I really needed it please
Answer
x.x.x.x.x = 5x
Step-by-step explanation:
![\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-3)^2+(y+7)^2=64\implies [x-\stackrel{h}{3}]^2+[y-(\stackrel{k}{-7})]^2=\stackrel{r}{8^2} \\\\\\ center~(3,-7)\qquad radius=8](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-3%29%5E2%2B%28y%2B7%29%5E2%3D64%5Cimplies%20%5Bx-%5Cstackrel%7Bh%7D%7B3%7D%5D%5E2%2B%5By-%28%5Cstackrel%7Bk%7D%7B-7%7D%29%5D%5E2%3D%5Cstackrel%7Br%7D%7B8%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Acenter~%283%2C-7%29%5Cqquad%20radius%3D8)
so, the broadcast location and range is more or less like the picture below.
Please, see the offered decision:
1) common equation for lines is y=kx+b. If k₁=k₂ (for line 1 and line 2) ⇒ 'line 1' || 'line 2'.
2) for line 3x+5y=6 k= -3/5. It means (according to item 1) for unknown line k is the same (-3/5).
3) using points (0;3) it is easy to find parameter b (x=0, y=3) via y=kx+b:
3=0*(-3/5)+b ⇔ b=3.
4) finaly (k=-3/5; b=3):