the quadrilateral ABCD on a coordinate plane has the following characteristics. AD can be represented by the equation y=-3x wher
1 answer:
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Answer:
See below
Step-by-step explanation:
Permutation is to select an object then arrange it and it cares about the orders while Combination is about only selecting an object without caring the orders.
Permutation can be expressed in math as:
where n is a number of total object and r is a number of selected object to arrange. Hence. n cannot be less than r.
Now let's see an example of permutation, suppose we have letter A, B and C. I'd like to know how many ways these words can be arranged:
Since there are 3 letters total and 3 selected letters to arrange then:
Therefore, there are 6 ways to arrange the letters - we can also demonstrate visually:
ABC - 1
ACB - 2
BAC - 3
BCA - 4
CAB - 5
CBA - 6
Notice that if you do visually, you'll get the same answer as the calculation of permutation!
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Combination can be expressed mathematically as:
The difference between permutation and combination is that you only find how many ways you can select object in combination. Therefore, no arrange and doesn't care about order, just ways to select.
Suppose we have same 3 letters: A, B and C. I want to find how many ways I can select these 3 letters:
Since there are 3 letters total and 3 selected letters:
Hence, there is only one way to select 3 letters. This makes sense because if you have 3 letters then you can only select 3 letters only one way.
Answer:
[a,b] divides n
Step-by-step explanation:
Let us denote the least common multiple of a and b [a,b]=m.
We want to prove that m divides n, where n is a multiple of a and b.
We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:
(1) ... n=mq+r, where 0<r<m
As n is a multiple of a and b, there exists s and t integers such that:
sa=n and tb=n
Same thing happens to m as it is the least common multiple, there exists u and v such that:
ua=m and vb=m
So (1) has the following form:
n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and
n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r
So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.
So this means that is and integer.
As m= vb, then is an integer, lets say
=v; and as m=ua, then
=u.
So v=
=a, so
divides a; on the other hand,
u=
=b, so
divides b. From this we can conclude that
is a common divisor of a and b.