Answer:
I will just give you the way of solving it
Step-by-step explanation:
First you have to change the sgape of the equation and it will become like this :
-6y=-2x+12
And then:
Y=0.33x+2
And then you have to give values to x and find y
For example i will make x zero (0)
It will become
Y=0.33(0)+12
=+12
And the points are (0,+12)
Go on like this and draw the line
Good luck
Sorry for bad english
Because the decimal is over 40 the answer is 15
Answer:
105³
Step-by-step explanation:
3x5=15
15x7=105
105 is your answer.
Answer: 13/3
Step-by-step explanation: TO turn a mixed number into an improper fraction, the best way is to multiply the denominator of the fraction by the whole number, then add the product to the numerator, then use that as the numerator of your new improper fraction, keeping the denominator the same.
>2 21/9
> 9*2 +21 =39
> 39/9
>13/3
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)