A cube because it has 6 faces/sides and 12 edges.
Answer:
![24\sqrt{5}\ yards](https://tex.z-dn.net/?f=24%5Csqrt%7B5%7D%5C%20yards)
Step-by-step explanation:
Let A1 be the area of one square and A2 be the area of second square
So,
A1 = s^2
where s is side of square
![s^2=125\\\sqrt{s^2}=\sqrt{125}\\s=\sqrt{25*5}\\ s= \sqrt{5^2 * 5}\\ s= 5\sqrt{5}](https://tex.z-dn.net/?f=s%5E2%3D125%5C%5C%5Csqrt%7Bs%5E2%7D%3D%5Csqrt%7B125%7D%5C%5Cs%3D%5Csqrt%7B25%2A5%7D%5C%5C%20s%3D%20%5Csqrt%7B5%5E2%20%2A%205%7D%5C%5C%20s%3D%205%5Csqrt%7B5%7D)
So side of one square is ![5\sqrt{5}](https://tex.z-dn.net/?f=5%5Csqrt%7B5%7D)
To calculate the length of fence we need to find the perimeter of the square
So,
P1 = 4 * s
![=4*5\sqrt{5} \\=20\sqrt{5}](https://tex.z-dn.net/?f=%3D4%2A5%5Csqrt%7B5%7D%20%5C%5C%3D20%5Csqrt%7B5%7D)
For second square:
![A_2=s^2\\5=s^2\\\sqrt{s^2}=5\\{s}=\sqrt{5}](https://tex.z-dn.net/?f=A_2%3Ds%5E2%5C%5C5%3Ds%5E2%5C%5C%5Csqrt%7Bs%5E2%7D%3D5%5C%5C%7Bs%7D%3D%5Csqrt%7B5%7D)
The perimeter will be:
![P_2 = 4*s\\=4 * \sqrt{5} \\=4\sqrt{5}](https://tex.z-dn.net/?f=P_2%20%3D%204%2As%5C%5C%3D4%20%2A%20%5Csqrt%7B5%7D%20%5C%5C%3D4%5Csqrt%7B5%7D)
So the total fence will be: P1+P2
![= 20\sqrt{5}+4\sqrt{5} \\= 24\sqrt{5}\ yards](https://tex.z-dn.net/?f=%3D%2020%5Csqrt%7B5%7D%2B4%5Csqrt%7B5%7D%20%5C%5C%3D%2024%5Csqrt%7B5%7D%5C%20yards)
Answer:
Step-by-step explanation:
a) prob(king first, Jack second)
= (4/52)(4/51) = 4/663
I don’t know about y but x=35 they are diagonal from one another so they will be the same