Okay so I got $742.50
This is what I did :
(1- .10) (825)
= 0.9 (825)
Which gives me $742.59
Hope this somehow helped
Check the picture below.
as we can see in the picture, the lengths for YZ and ZX are very straight, let's get the side YX
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ Y(\stackrel{x_1}{0}~,~\stackrel{y_1}{-2})\qquad X(\stackrel{x_2}{2}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ YX=\sqrt{[2 - 0]^2 + [4 - (-2)]^2}\implies YX=\sqrt{2^2+(4+2)^2} \\\\\\ YX=\sqrt{4+6^2}\implies YX=\sqrt{40} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{2~~+~~6~~+~~\sqrt{40}~~\approx~~14.32} ~\hfill \stackrel{\textit{\Large Area}}{\cfrac{1}{2}(2)(6)\implies 6}](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20Y%28%5Cstackrel%7Bx_1%7D%7B0%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20X%28%5Cstackrel%7Bx_2%7D%7B2%7D~%2C~%5Cstackrel%7By_2%7D%7B4%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20YX%3D%5Csqrt%7B%5B2%20-%200%5D%5E2%20%2B%20%5B4%20-%20%28-2%29%5D%5E2%7D%5Cimplies%20YX%3D%5Csqrt%7B2%5E2%2B%284%2B2%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20YX%3D%5Csqrt%7B4%2B6%5E2%7D%5Cimplies%20YX%3D%5Csqrt%7B40%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Perimeter%7D%7D%7B2~~%2B~~6~~%2B~~%5Csqrt%7B40%7D~~%5Capprox~~14.32%7D%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Area%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%282%29%286%29%5Cimplies%206%7D)
Step-by-step explanation:
1. 10-(2+3) = 5
2. 2x3 - 10 = -4
3. 3 x (4-2) = 6
4. (3+5) / 4 = 2
5. (5-2) x 3 = 9
6. (9-4) + 5 = 10
Answer:
384cm^2,
Step-by-step explanation:
The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 384 cm^2,
