well for a line, to get its slope all we need is two points, so let's use (-6, 5) and (0, 1), and get the equation of it.
![\bf \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-5=-\cfrac{5}{6}[x-(-6)]\implies y-5=-\cfrac{5}{6}(x+6) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \textit{to get the x-intercept, we set y = 0, solve for \underline{x}} \\\\\\ 0-5=-\cfrac{5}{6}(x+6)\implies -30=5x+30\implies -60=5x \\\\\\ \cfrac{-60}{5}=x\implies -12=x~\hfill \boxed{\stackrel{x-intercept}{(-12,0)}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%0A%5Ccline%7B1-1%7D%0A%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%0A%5Ccline%7B1-1%7D%0A%5C%5C%0Ay-y_1%3Dm%28x-x_1%29%0A%5C%5C%5C%5C%0A%5Ccline%7B1-1%7D%0A%5Cend%7Barray%7D%5Cimplies%20y-5%3D-%5Ccfrac%7B5%7D%7B6%7D%5Bx-%28-6%29%5D%5Cimplies%20y-5%3D-%5Ccfrac%7B5%7D%7B6%7D%28x%2B6%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ctextit%7Bto%20get%20the%20x-intercept%2C%20we%20set%20y%20%3D%200%2C%20solve%20for%20%5Cunderline%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A0-5%3D-%5Ccfrac%7B5%7D%7B6%7D%28x%2B6%29%5Cimplies%20-30%3D5x%2B30%5Cimplies%20-60%3D5x%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B-60%7D%7B5%7D%3Dx%5Cimplies%20-12%3Dx~%5Chfill%20%5Cboxed%7B%5Cstackrel%7Bx-intercept%7D%7B%28-12%2C0%29%7D%7D)
now, where's the y-intercept of that line? well, to get the y-intercept, we set x = 0 and solve for "y"....hmmmm wait a second, notice (0, 1), x = 0, y = 1, that's the y-intercept already.
Answer:
Step-by-step explanation:
Before solving this we have to know that,
(-) ×(-)=(+)
(+)×(+) =(+)
(+)×(-)=(-)
Lets solve now,
Hope this helps you
Let me know if you have any other questions :-)
Answer:
cccccccdddffff
Step-by-step explanation:
Answer:
1.5333333
Step-by-step explanation:
7 1/3 + 5 4/5 = 23/15 =1 8/15 =1.53
Answer:
<h2>2.7777777778</h2>
Step-by-step explanation:
