to get the equation of any straight line, we simply need two points off of it, so for the equation in the table, let'use the points in the picture below
![(\stackrel{x_1}{1}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-2)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{2+2}{2}\implies 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}{(-2)}=\stackrel{m}{2}(x-\stackrel{x_1}{1})\implies y+2=2x-2](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B2%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B2%7D-%5Cstackrel%7By1%7D%7B%28-2%29%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B3%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B2%2B2%7D%7B2%7D%5Cimplies%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%7B%28-2%29%7D%3D%5Cstackrel%7Bm%7D%7B2%7D%28x-%5Cstackrel%7Bx_1%7D%7B1%7D%29%5Cimplies%20y%2B2%3D2x-2)
![y=2x\stackrel{\stackrel{b}{\downarrow }}{-4}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}](https://tex.z-dn.net/?f=y%3D2x%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B-4%7D%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D)
so the equation in the table has a y-intercept of -4.
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
![y+\cfrac{1}{4}x=2\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{4}} x+2\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%2B%5Ccfrac%7B1%7D%7B4%7Dx%3D2%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B1%7D%7B4%7D%7D%20x%2B2%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 hmmm we're really looking for an equation whose slope is -1/4 and with a y-intercept greater than -4, hmmm let's recall something, that on the negative side of the number line, the closer to 0, the greater.
that said, on the negative side of the number line -10 is much larger than -1,000,000, because -10 is closer than -1,000,000 to 0, and so on.
so hmmm what's greater than -4? hmm well, heck let's use -2
![y = -\cfrac{1}{4}x - 2](https://tex.z-dn.net/?f=y%20%3D%20-%5Ccfrac%7B1%7D%7B4%7Dx%20-%202)
Answer:
go on tigermath
Step-by-step explanation:
I hope this helps. If you are still confused just tell me and I’ll explain it again
Answer:
Lm
Step-by-step explanation:
average = (sum of numbers)/(number of numbers)
Solve the equation for "sum of numbers":
sum of numbers = average * number of numbers
sum of numbers = m * L
sum of numbers = Lm
Answer: B) Lm
Answer:
31/100 and 215/1,000 and 6704/10,000
the pattern is you are adding another 0 to the denominator