to get the equation of any straight line, we simply need two points off of it, let's use the points from the picture below.
![(\stackrel{x_1}{8}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{12}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{12}-\underset{x_1}{8}}}\implies \cfrac{2}{4}\implies \cfrac{1}{2}](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B8%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B12%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B5%7D-%5Cstackrel%7By1%7D%7B3%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B12%7D-%5Cunderset%7Bx_1%7D%7B8%7D%7D%7D%5Cimplies%20%5Ccfrac%7B2%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D)
![\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}{3}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{8}) \\\\\\ y-3=\cfrac{1}{2}x-4\implies y=\cfrac{1}{2}x-1](https://tex.z-dn.net/?f=%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%7B3%7D%3D%5Cstackrel%7Bm%7D%7B%5Ccfrac%7B1%7D%7B2%7D%7D%28x-%5Cstackrel%7Bx_1%7D%7B8%7D%29%20%5C%5C%5C%5C%5C%5C%20y-3%3D%5Ccfrac%7B1%7D%7B2%7Dx-4%5Cimplies%20y%3D%5Ccfrac%7B1%7D%7B2%7Dx-1)
if we already have the slope, and we can see the y-intercept on the table, then we can simply use the slopel-intercept form and plug both of them in.
![\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \begin{array}{|c|ll} \cline{1-1} slope\\ \cline{1-1} \cfrac{1}{2}\\ \cline{1-1} y-intercept&\\\cline{1-1} (0~~,~~-1)\\ \cline{1-1} \end{array}~\hfill y=\cfrac{1}{2}x-1](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cqquad%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope%5C%5C%20%5Ccline%7B1-1%7D%20%5Ccfrac%7B1%7D%7B2%7D%5C%5C%20%5Ccline%7B1-1%7D%20y-intercept%26%5C%5C%5Ccline%7B1-1%7D%20%280~~%2C~~-1%29%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D~%5Chfill%20y%3D%5Ccfrac%7B1%7D%7B2%7Dx-1)
Answer:
C = 18.1°
Step-by-step explanation:
A = 142 m²
a = 24 cm, b = 38 cm
Use the formula A = a*b*sinC/2
142 = 24 * 38 * sinC/2 = 456*sinC
sinC = 0.3114
C = 18.1°
Attached a solution and showed work.
Answer:3 hours and 30 minutes
Answer:
Step-by-step explanation:
Radius of the circle
Center of the circle (h, k) = (1.3, - 3.5)
Equation of the circle in center radius form is given as:
Plugging the values of h, k and r in the above equation, we find:
This is the required equation of the circle.