Answer:
c,d
Step-by-step explanation:
![ab^{-3x}=a(b^{-3})^x=a(\frac{1}{b^3} )^x=a[(\frac{1}{b} )^3]^x\\=a(\frac{1}{b} )^{3x}](https://tex.z-dn.net/?f=ab%5E%7B-3x%7D%3Da%28b%5E%7B-3%7D%29%5Ex%3Da%28%5Cfrac%7B1%7D%7Bb%5E3%7D%20%29%5Ex%3Da%5B%28%5Cfrac%7B1%7D%7Bb%7D%20%29%5E3%5D%5Ex%5C%5C%3Da%28%5Cfrac%7B1%7D%7Bb%7D%20%29%5E%7B3x%7D)
Answer:
The inequality that represents the age of the group, "x", is: 
Step-by-step explanation:
To express this problem in an inequality we will attribute the age of the members on the group with the variable "x". There are two available information about "x", the first states that every member of the group is older than 17 years, therefore we can create a inequality based on that:

While the second data from the problem states that none of than is older than 54 years old, this implies that they can be at most that old, therefore the inequality that represents this is:

In order for both to be valid at the same time x must be greater than 17 and less or equal to 54, therefore we finally have:

Answer:
if the question is (6x+4y) - 2y
6x+2y or 3x+y
but if its (6x+4y)(-2y)
-6xy+(-8y)
Step-by-step explanation:
In order to use the elimination method, we have to multiply the equation by some number, so that one of the variable has the same coefficient.
For example, multiplying the first equation by 3 and the second by 2 gives the following, equivalent system:

Now, we can subtract the two equations, and we will cancel (eliminate) the x variable:

Now that y is known, plug it into one of the equations: for example, if we use the first one we get

There are 120 ways in which 5 riders and 5 horses can be arranged.
We have,
5 riders and 5 horses,
Now,
We know that,
Now,
Using the arrangement formula of Permutation,
i.e.
The total number of ways
,
So,
For n = 5,
And,
r = 5
As we have,
n = r,
So,
Now,
Using the above-mentioned formula of arrangement,
i.e.
The total number of ways
,
Now,
Substituting values,
We get,

We get,
The total number of ways of arrangement = 5! = 5 × 4 × 3 × 2 × 1 = 120,
So,
There are 120 ways to arrange horses for riders.
Hence we can say that there are 120 ways in which 5 riders and 5 horses can be arranged.
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