now, we're assuming they're directly proportional, so say y = miles, x = hours.
![\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \textit{we know that } \begin{cases} y=\frac{4}{5}\\[0.8em] x=\frac{1}{8} \end{cases}\implies \cfrac{4}{5}=k\cfrac{1}{8}\implies \cfrac{4}{5}=\cfrac{k}{8} \\\\\\ \cfrac{8\cdot 4}{5}=k\implies \cfrac{32}{5}=k\implies 6\frac{2}{5}=k](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bdirect%20proportional%20variation%7D%20%5C%5C%5C%5C%20%5Ctextit%7B%5Cunderline%7By%7D%20varies%20directly%20with%20%5Cunderline%7Bx%7D%7D%5Cqquad%20%5Cqquad%20y%3Dkx%5Cimpliedby%20%5Cbegin%7Barray%7D%7Bllll%7D%20k%3Dconstant%5C%20of%5C%5C%20%5Cqquad%20variation%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Ctextit%7Bwe%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20y%3D%5Cfrac%7B4%7D%7B5%7D%5C%5C%5B0.8em%5D%20x%3D%5Cfrac%7B1%7D%7B8%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B5%7D%3Dk%5Ccfrac%7B1%7D%7B8%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B5%7D%3D%5Ccfrac%7Bk%7D%7B8%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B8%5Ccdot%204%7D%7B5%7D%3Dk%5Cimplies%20%5Ccfrac%7B32%7D%7B5%7D%3Dk%5Cimplies%206%5Cfrac%7B2%7D%7B5%7D%3Dk)
Step 1
Given;

Required; Find the equation of a line that goes through the point ( 1,- 6 ) and is perpendicular to the line: y = (1 / 8)x - 9
Step 2
For perpendicular lines, the relationship between the slopes is;


The y-intercept is found thus;

Answer; The required equation will be;
Answer:
1. (x+2y)^2-ab x^2+4xy+4y^2-ab
2. 3n-13<9 n<22/3
3. 6+n/7=4 n/7=-2
Answer:
33, 36, 39, 40
Step-by-step explanation: