V2 + 2v + 1 is the product of which two binomials?

2 answers:

8 0

The answer is (v+1)^2

(side note: when writing exponents use "^" to indicate the following number is an exponent. ex. 3 squared would be 3^2)

juin [17]3 years ago

6 0

(V+1)2 is that right

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Why is 6/11=x/33 not a proportion

sertanlavr [38]

6/11=x/33 is not a proportion because they aren't equal to each other.

6 0

3 years ago

What is Permutations and Combinations?

marissa [1.9K]

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:

$\displaystyle{_n P _r = \dfrac{n!}{(n-r)!} \ \ \ (n \geq r) }$

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:

$\displaystyle{_3 P _3 = \dfrac{3!}{(3-3)!}}\\\\\displaystyle{_3 P _3 = \dfrac{3 \times 2 \times 1}{0!}}\\\\\displaystyle{_3 P _3 = \dfrac{6}{1}}\\\\\displaystyle{_3 P _3 = 6}$

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!

----

Combination can be expressed mathematically as:

$\displaystyle{_n C _r = \dfrac{n!}{(n-r)!r!} = \dfrac{_n P _r}{r!} \ \ \ (n \geq r) }$

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:

$\displaystyle{_3 C _3 = \dfrac{3!}{(3-3)!3!}}\\\\\displaystyle{_3 C _3 = \dfrac{3!}{0!3!}}\\\\\displaystyle{_3 C _3 = \dfrac{3!}{3!}}\\\\\displaystyle{_3 C _3 = 1}$