5x/3=35
5x = 35*3=105
x= 105/5 = 21
double check 5 x 21 = 105
105/3 = 35
To solve this problem you must apply the proccedure shown below:
1. You have that:
- T<span>he equation d=1/2(n(n-3)) gives the number of diagonals for the polygon.
- The polygon that has 65 diagonals..
2. When you clear n, you obtain:
d=n(n-3)/2
d=(n^2-3n)/2
2x65=n^2-3n
n^2-3n-130=0
3. When you solve the quadratic equation, you obtain:
n=13
Therefore, the answer is: 13 sides. </span>
Answer:
1. 32
2. 170
3. 29
Step-by-step explanation:
1. is just 10 divided by 320, which gives you 32
2. is 10 times 17 which is 170
3. you need 29 tens to get 290.
Answer:
The statement is true is for any 
.
Step-by-step explanation:
First, we check the identity for 
:
The statement is true for .
Then, we have to check that identity is true for 
, under the assumption that 
is true:
![(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](https://tex.z-dn.net/?f=%281%5E%7B2%7D%2B2%5E%7B2%7D%2B3%5E%7B2%7D%2B...%2Bk%5E%7B2%7D%29%20%2B%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5E%7B2%7D%20%3D%20%5Cfrac%7B%28k%2B1%29%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29%2B1%5D%7D%7B3%7D)
![\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ccdot%20%282%5Ccdot%20k%20-1%29%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%7D%7B3%7D%20%2B%5B2%5Ccdot%20%28k%2B1%29-1%5D%5E%7B2%7D%20%3D%20%5Cfrac%7B%28k%2B1%29%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29%2B1%5D%7D%7B3%7D)
![\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ccdot%20%282%5Ccdot%20k%20-1%29%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%2B3%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5E%7B2%7D%7D%7B3%7D%20%3D%20%5Cfrac%7B%28k%2B1%29%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29%2B1%5D%7D%7B3%7D)
![(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)](https://tex.z-dn.net/?f=%282%5Ccdot%20k%20%2B1%29%5Ccdot%20%5Bk%5Ccdot%20%282%5Ccdot%20k%20-1%29%2B3%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%5D%20%3D%20%28k%2B1%29%20%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%5Ccdot%20%282%5Ccdot%20k%20%2B3%29)
Therefore, the statement is true for any .
