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3 years ago

Without his pencil or calculator, Joey knows that 2x³+3x²− 1 = 0 has at least one real solution. How does he know?

Mathematics

1 answer:

vova2212 [387]3 years ago

6 0

Answer: Find the discriminant

Check explanation

Step-by-step explanation:

b^2-4ac

if it is equal to 0 there is no solution

Greater then 0 there is 2 real solution

Less then 0 there is 1 real

3^2-4(2*-1)

9-4(-2)

9(8)

72 there is at least one real solution

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Step-by-step explanation:

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3 years ago

Type the correct answer in the box.

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V should be written as (1/3) pi r^2 h

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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:

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

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.

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2 years ago