Check the picture below.
now, you can pretty much count the units off the grid for the segments ST and RU, so each is 7 units long, and are parallel, meaning that the other two segments are also parallel, and therefore the same length each.
so we can just find the length for hmmmm say SR, since SR = TU, TU is the same length,
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ S(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad R(\stackrel{x_2}{-5}~,~\stackrel{y_2}{5})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ SR=\sqrt{[-5-(-2)]^2+[5-1]^2}\implies SR=\sqrt{(-5+2)^2+(5-1)^2} \\\\\\ SR=\sqrt{(-3)^2+4^2}\implies SR=\sqrt{25}\implies SR=5](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AS%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20%0AR%28%5Cstackrel%7Bx_2%7D%7B-5%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ASR%3D%5Csqrt%7B%5B-5-%28-2%29%5D%5E2%2B%5B5-1%5D%5E2%7D%5Cimplies%20SR%3D%5Csqrt%7B%28-5%2B2%29%5E2%2B%285-1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ASR%3D%5Csqrt%7B%28-3%29%5E2%2B4%5E2%7D%5Cimplies%20SR%3D%5Csqrt%7B25%7D%5Cimplies%20SR%3D5)
sum all segments up, and that's perimeter.
Answer:
Glass A holds the most, It holds more than glass B because of its cone shape.
Step-by-step explanation:
I hope I helped you, now have a good day and good luck!! ;)
Answer:
1 in 47.
Step-by-step explanation:
Add all of the outcomes (28 + 2 + 3 + 14).
You'll get 47.
Do not add the 1s.
There is your answer.
1 in 47.
<span>1st piece: x feet
2nd piece: 8-x feet
--------------------------
Use the "x" piece to form a circle:
Circumference = "x".
2(pi)*radius = x
radius = x/(2pi)
So, Area = pi[x/(2)]^2 = x^2/(4pi) = (1/4pi)x^2
=========================================================
Use the 10-x piece to form square:
side = (1/4pi)(10-x)
Area = side^2 = (1/16)(10-x)^2
Hope this helps! :)</span>