All of these are mostly approximately
1) 250.3
2)3421.19
3)11494.04
4) 3.108
Answer:
x = 47
Step-by-step explanation:
86° and 2x° are supplementary angles, which means that added together they should equal 180° (which is a line). Aka
86 + 2x = 180 (Let's solve for x) (subtract both sides by 86 to get x on one side)
2x = 94 (divide both sides by 2 to get x by itself)
x = 47
Answer: 306 inches or 25.5 feet or 8.5 yards.
Step-by-step explanation:
Given: Carol has 90 inches of ribbon, Tino has 7.5 feet of ribbon, and Baxter has 3.5 yards of ribbon.
1 feet = 12 inches
length of ribbon Tino has = 12 x 7.5 = 90 inches
1 yard = 36 inches
length of ribbon Baxter has = 3.5 x 36 = 126 inches
Total ribbon they have = (length of ribbon Carol has) + (length of ribbon Tino has) + (length of ribbon Baxter has)
= (90+90+126) inches
= 306 inches
In feet , Total ribbon =
![[\text{ 1 inch}=\dfrac{1}{12}\text{ feet}]](https://tex.z-dn.net/?f=%5B%5Ctext%7B%201%20inch%7D%3D%5Cdfrac%7B1%7D%7B12%7D%5Ctext%7B%20feet%7D%5D)

In yards, Total ribbon =
![[\text{1 feet}=\dfrac{1}{3}\text{ yard}]](https://tex.z-dn.net/?f=%5B%5Ctext%7B1%20feet%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Ctext%7B%20yard%7D%5D)

Hence, the total length of ribbon the three friends have is 306 inches or 25.5 feet or 8.5 yards.
Answer:
The number of possible choices of my team and the opponents team is

Step-by-step explanation:
selecting the first team from n people we have
possibility and choosing second team from the rest of n-1 people we have 
As { A, B} = {B , A}
Therefore, the total possibility is 
Since our choices are allowed to overlap, the second team is 
Possibility of choosing both teams will be
![\frac{n(n-1)}{2} * \frac{n(n-1)}{2} \\\\= [\frac{n(n-1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%20%20%2A%20%20%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%20%20%5C%5C%5C%5C%3D%20%5B%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
We now have the formula
1³ + 2³ + ........... + n³ =![[\frac{n(n+1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
1³ + 2³ + ............ + (n-1)³ = ![[x^{2} \frac{n(n-1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5Bx%5E%7B2%7D%20%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
=![\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] = [\frac{n(n-1)}{2}]^{3}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dn-1%5C%5CE%5C%5Ci%3D1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5B%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%5E%7B3%7D)