Answer:
NO because 7+5=12
The sum must be larger (not equal to) the longest side. If it's exactly the same, there's no room to for the shorter two sides to angle up to form a triangle.
Answer:
C. 1/3<b<3/2
D. 1/6<x<2
Step-by-step explanation:
Hope this helps
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)
Answer:
Im pretty sure the answer is B, forgive me if I am wrong.
Step-by-step explanation:
ANSWER: 8/21
Step:
1. Find the GCF (Greatest Common Factor)
The two denominator which is the bottom number have to be the same ALWAYS when adding and subtracting the fraction. So to find the GCF, you keep multiplying until you see the same number, only use the bottom number to find the GCF
7: 7, 14, 21, 28
3: 3, 6, 9, 12, 15, 18, 21, 24, 27
GCF= 21
2. 5/7 -1/3
Multiple both to get 21
5/7 x 3/3 = 15/21
1/3 x 7/7 = 7/21
3. Sove
15/21 - 7/21
= 8/21
