bearing 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 (\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-2}{2-(-3)}\implies \cfrac{1-2}{2+3}\implies -\cfrac{1}{5} \\\\\\ \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-2=-\cfrac{1}{5}[x-(-3)]\implies y-2=-\cfrac{1}{5}(x+3)](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B2%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B2%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20m%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B1-2%7D%7B2-%28-3%29%7D%5Cimplies%20%5Ccfrac%7B1-2%7D%7B2%2B3%7D%5Cimplies%20-%5Ccfrac%7B1%7D%7B5%7D%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-2%3D-%5Ccfrac%7B1%7D%7B5%7D%5Bx-%28-3%29%5D%5Cimplies%20y-2%3D-%5Ccfrac%7B1%7D%7B5%7D%28x%2B3%29)
This question can be found using the distance formula. This formula can be written as d=sqrt((x2-x1)2+(y2-y1)2). I’d advise looking up the formula, as mobile wont let me format the exponents. Plug in your values, and you should get d=sqrt(3481+3136). Simplify this, and you will get d=sqrt(6617).
Answer:
Find the 58th term of the arithmetic sequence
−20
Step-by-step explanation:
True, it has a 90 degree angle. It’s clear as day. And the Pythagorean theorem works perfectly fine with the two legs and hypotenuse
