Its like j have to add multilp and all that
keeping in mind that standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
![\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-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{6}{7}}[x-\stackrel{x_1}{(-3)}]\implies y+1=\cfrac{6}{7}(x+3) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{7}}{7(y+1)=7\left( \cfrac{6}{7}(x+3) \right)}\implies 7y+7=6(x+3)\implies 7y+7=6x+18 \\\\\\ 7y=6x+11\implies -6x+7y=11\implies 6x-7y=-11](https://tex.z-dn.net/?f=%5Cbf%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%28-1%29%7D%3D%5Cstackrel%7Bm%7D%7B%5Ccfrac%7B6%7D%7B7%7D%7D%5Bx-%5Cstackrel%7Bx_1%7D%7B%28-3%29%7D%5D%5Cimplies%20y%2B1%3D%5Ccfrac%7B6%7D%7B7%7D%28x%2B3%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B7%7D%7D%7B7%28y%2B1%29%3D7%5Cleft%28%20%5Ccfrac%7B6%7D%7B7%7D%28x%2B3%29%20%5Cright%29%7D%5Cimplies%207y%2B7%3D6%28x%2B3%29%5Cimplies%207y%2B7%3D6x%2B18%20%5C%5C%5C%5C%5C%5C%207y%3D6x%2B11%5Cimplies%20-6x%2B7y%3D11%5Cimplies%206x-7y%3D-11)
Answer:
I need to see the data set and graph to be able to help you more
Step-by-step explanation:
Answer:
5/39
Step-by-step explanation:
10divided by 78 is 10/78.
you can simplify it to 5/39
this is the simplest form
The vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)
<h3>How to determine the new vertices?</h3>
The vertices are given as:
a (-6,-7)
b(-3,-10)
c(-5,2)
The transition rule is given as
(x,y)-->(x,y-3)
So, we have
a' = (-6, -7 - 3)
a' = (-6, -10)
b' = (-3, -10 - 3)
b' = (-3, -13)
c' = (-5, 2 - 3)
c' = (-5, -1)
Hence, the vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)
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brainly.com/question/4289712
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