Answer:
Hello again!
Original coordinates: 5, -2
New coordinates: 0, 4
Heres a picture demonstration to find the new coordinates. How I moved 5 left and 6 up
Hope that helps!
X = 47*
Explanation:
So to find x you first find 2x = ?
86* + 2x* = 180
180* - 86* = 94*
2x* = 94*
94/2 = 47*
X = 47*
![\bf \begin{cases} t=tacos\\ m=milk \end{cases}~\hspace{7em} \begin{cases} \stackrel{1~taco}{t}+\stackrel{1~glass}{m}=\stackrel{\$}{2.10}\\[0.8em] \stackrel{2~tacos}{2t}+\stackrel{3~glasses}{3m}=\stackrel{\$}{5.15} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{array}{llll} t+m=2.10&\stackrel{multiplied~by}{\times -3}\implies &\stackrel{notice}{-3t\stackrel{\downarrow }{-3m}=-6.3}\\[0.8em] 2t+3m=5.15&\implies &~~2t+3m=5.15\\ \cline{3-3} &&-t+0m=-1.15 \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0At%3Dtacos%5C%5C%0Am%3Dmilk%0A%5Cend%7Bcases%7D~%5Chspace%7B7em%7D%0A%5Cbegin%7Bcases%7D%0A%5Cstackrel%7B1~taco%7D%7Bt%7D%2B%5Cstackrel%7B1~glass%7D%7Bm%7D%3D%5Cstackrel%7B%5C%24%7D%7B2.10%7D%5C%5C%5B0.8em%5D%0A%5Cstackrel%7B2~tacos%7D%7B2t%7D%2B%5Cstackrel%7B3~glasses%7D%7B3m%7D%3D%5Cstackrel%7B%5C%24%7D%7B5.15%7D%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0At%2Bm%3D2.10%26%5Cstackrel%7Bmultiplied~by%7D%7B%5Ctimes%20-3%7D%5Cimplies%20%26%5Cstackrel%7Bnotice%7D%7B-3t%5Cstackrel%7B%5Cdownarrow%20%7D%7B-3m%7D%3D-6.3%7D%5C%5C%5B0.8em%5D%0A2t%2B3m%3D5.15%26%5Cimplies%20%26~~2t%2B3m%3D5.15%5C%5C%0A%5Ccline%7B3-3%7D%0A%26%26-t%2B0m%3D-1.15%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill)
![\bf -t=-1.15\implies t=\cfrac{-1.15}{-1}\implies \blacktriangleright t=1.15 \blacktriangleleft \\\\\\ \stackrel{\textit{and now substituting on the 2nd equation}}{2(1.15)+3m=5.15}\implies 2.3+3m=5.15 \\\\\\ 3m=2.85\implies m=\cfrac{2.85}{3}\implies \blacktriangleright m=0.95 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20-t%3D-1.15%5Cimplies%20t%3D%5Ccfrac%7B-1.15%7D%7B-1%7D%5Cimplies%20%5Cblacktriangleright%20t%3D1.15%20%5Cblacktriangleleft%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7B%5Ctextit%7Band%20now%20substituting%20on%20the%202nd%20equation%7D%7D%7B2%281.15%29%2B3m%3D5.15%7D%5Cimplies%202.3%2B3m%3D5.15%0A%5C%5C%5C%5C%5C%5C%0A3m%3D2.85%5Cimplies%20m%3D%5Ccfrac%7B2.85%7D%7B3%7D%5Cimplies%20%5Cblacktriangleright%20m%3D0.95%20%5Cblacktriangleleft)
notice, we used -3 in the multiplication to <u>eliminate</u> the m's.
Step 1: recognize the geometry as a 6 m square and 1/4 of a circle of radius 6 m.
Step 2: recall the formula for the area of a square is
A = s²
so the area of the square portion of the geometry is
A = (6 m)² = 36 m²
Step 3: recall the formula for the area of a circle is
A = πr²
so the area of the 1/4 circle will be
A = (1/4)*π*(6 m)² = 9π m²
Step 4: add the areas of the parts to find the area of the whole.
whole area = (36 m²) +(9π m²)
whole area = (36 +9π) m²
whole area ≈ 64.2743 m²