Answer:
The options are not shown, so i will answer in a general way.
Let's define the variables:
h = number of hats
m = number of mugs.
We know that a total of 1000 items were ordered, then:
h + m = 1000
We also know that we have 3 times more mugs than hats, this can be written as:
m = 3*h
Now we have the system of equations:
h + m = 1000
m = 3*h
To solve these, we usually start by isolating one of the variables in one equation and then replace that in the other equation, but in this case, we already have m isolated in the second equation, then we can replace that in the first equation to get:
h + m = 1000
h + (3*h) = 1000
Now we can solve this equation for h, and find the number of hats ordered.
4*h = 1000
h = 1000/4 = 250
There were 250 hats ordered.
Answer:
7(x)=5x+7
to find the x / zero intercept, plug in f (x) = 0
0=5x+7
move the left side and change the sign
-5x7
divide both of the equation by -5
x= 
same steps for g(x)=-2x-4
x=-2
It's the first one. y = 3x
proportional means they are just some multiple of the other.
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)
X=-1.8
To solve first use the distributive property to change the equation to -2x+1 1/3+1=5
Then you subtract 1 from 5 so then the equation looks like this: -2x+1 1/3=4
Next subtract 1 1/3 from 4 to get 3.6
then divide it by -2 and you get -1.8