since you already know how to get the slopes and equation, let's do the equations of each of these two without much fuss.
then we'll have two equations, namely a <u>system of equations of two variables</u>, and we'll solve it by <u>substitution</u>.
first equation is
![\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{10-14}{6-(-2)}\implies \cfrac{-4}{6+2}\implies -\cfrac{1}{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-14=-\cfrac{1}{2}[x-(-2)] \\\\\\ y-14=-\cfrac{1}{2}(x+2)\implies y=-\cfrac{1}{2}(x+2)+14](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B14%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B10%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%7B10-14%7D%7B6-%28-2%29%7D%5Cimplies%20%5Ccfrac%7B-4%7D%7B6%2B2%7D%5Cimplies%20-%5Ccfrac%7B1%7D%7B2%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-14%3D-%5Ccfrac%7B1%7D%7B2%7D%5Bx-%28-2%29%5D%20%5C%5C%5C%5C%5C%5C%20y-14%3D-%5Ccfrac%7B1%7D%7B2%7D%28x%2B2%29%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B2%7D%28x%2B2%29%2B14)
second equation is
![\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-(-4)}{2-(-1)}\implies \cfrac{5+4}{2+1}\implies 3 \\\\\\ \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-(-4)=3[x-(-1)] \\\\\\ y+4=3(x+1)\implies y=3(x+1)-4](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B2%7D~%2C~%5Cstackrel%7By_2%7D%7B5%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%7B5-%28-4%29%7D%7B2-%28-1%29%7D%5Cimplies%20%5Ccfrac%7B5%2B4%7D%7B2%2B1%7D%5Cimplies%203%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-%28-4%29%3D3%5Bx-%28-1%29%5D%20%5C%5C%5C%5C%5C%5C%20y%2B4%3D3%28x%2B1%29%5Cimplies%20y%3D3%28x%2B1%29-4)
now, let's recall, let's substitute "y" in the second equation,
![\bf \stackrel{\stackrel{y}{\downarrow }}{-\cfrac{1}{2}(x+2)+14}=3(x+1)-4\implies -\cfrac{x}{2}-1+14=3x+3-4 \\\\\\ -\cfrac{x}{2}+14=3x\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{-x+28=6x} \\\\\\ 28=7x\implies \cfrac{28}{7}=x\implies \blacktriangleright 4=x \blacktriangleleft \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cstackrel%7By%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B1%7D%7B2%7D%28x%2B2%29%2B14%7D%3D3%28x%2B1%29-4%5Cimplies%20-%5Ccfrac%7Bx%7D%7B2%7D-1%2B14%3D3x%2B3-4%20%5C%5C%5C%5C%5C%5C%20-%5Ccfrac%7Bx%7D%7B2%7D%2B14%3D3x%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B2%7D%7D%7B-x%2B28%3D6x%7D%20%5C%5C%5C%5C%5C%5C%2028%3D7x%5Cimplies%20%5Ccfrac%7B28%7D%7B7%7D%3Dx%5Cimplies%20%5Cblacktriangleright%204%3Dx%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \stackrel{\textit{now, let's substitute the found \underline{x} in the 2nd equation}}{y=3(4+1)-4\implies y=3(5)-4}\implies \blacktriangleright y= 11 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (4,11)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bnow%2C%20let%27s%20substitute%20the%20found%20%5Cunderline%7Bx%7D%20in%20the%202nd%20equation%7D%7D%7By%3D3%284%2B1%29-4%5Cimplies%20y%3D3%285%29-4%7D%5Cimplies%20%5Cblacktriangleright%20y%3D%2011%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%284%2C11%29~%5Chfill)
That depends on what the digits are. It could be anywhere between zero times and infinite times. If both are the same digit and not zero, then the tenth is ten times the size of the hundredth.
Answer:
The missing length is 7
Step-by-step explanation:
Answer:
y = 7.5![\sqrt{x}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D)
Step-by-step explanation:
given y is directly proportional to
then the equation relating them is
y = k
← k is the constant of proportion
to find k use the condition y = 75 , x = 100 , then
75 = k
= 10k ( divide both sides by 10 )
7.5 = k
y = 7.5
← equation of proportion