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)
I don’t know u tell me that’s the answer
I believe it’s A! because they would be wanting to control their new subjects that’s where dominating comes in
Answer:
x = 9
Step-by-step explanation:
for any two intersecting chords, the product of their section lengths will be identical.
20(2x + 3) = 14(3x + 3)
40x + 60 = 42x = 42
18 = 2x
x = 9
Answer:
k=2
Step-by-step explanation:
Open parenthesis:
6k-12=3k-6
Put all like terms on one side:
6k-3k=-6+12
Simplify:
3k=6
k=2
<em>~Stay golden~ :)</em>