129 FT of fencing is required for the problem
Graph b would be my guess!
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
![y-1=\stackrel{\stackrel{m}{\downarrow }}{2}(x-4)\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}](https://tex.z-dn.net/?f=y-1%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B2%7D%28x-4%29%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D)
so we're really looking for the equation of a line with a slope of 2 and that it passes through (4 , 1)
![(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{2}(x-\stackrel{x_1}{4})](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B4%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Chspace%7B10em%7D%20%5Cstackrel%7Bslope%7D%7Bm%7D%20~%3D~%202%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B1%7D%3D%5Cstackrel%7Bm%7D%7B2%7D%28x-%5Cstackrel%7Bx_1%7D%7B4%7D%29)
kinda twins.
Answer:
![x\leq \$15](https://tex.z-dn.net/?f=x%5Cleq%20%5C%2415)
Step-by-step explanation:
Let
x------> amount of money that Samantha spends weekly on her school lunch
we know that
The inequality that represent this situation is
![x\leq \$15](https://tex.z-dn.net/?f=x%5Cleq%20%5C%2415)
All positive real number less than or equal to 15
Answer:
122
Step-by-step explanation:
1^6 = 1
11^2 = 121
1 + 121 = 122
<em>hope this helps</em>