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Oxana [17]
3 years ago
6

Write 25*24*23*22 as a quotient of factorials and as a permutation

Mathematics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

(a)

25\times 24\times 23\times 22=\frac{25!}{21!}

(b)

25\times 24\times 23\times 22=P(25,4)

Step-by-step explanation:

Quotient of factorials:

we are given

25\times 24\times 23\times 22

we can multiply top and bottom term by 21!

25\times 24\times 23\times 22=\frac{25\times 24\times 23\times 22\times 21!}{21!}

we can write as

25\times 24\times 23\times 22=\frac{25!}{21!}

As a permutation:

we know permutation formula

P(n,r)=\frac{n!}{(n-r)!}

now, we can compare and find 'n' and 'r'

n=25

n-r=21

we can plug back n=25

25-r=21

r=4

so, we can write

25\times 24\times 23\times 22=P(25,4)

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2 years ago
6. Solve by factoring: 2x²-11x+14=0
Mariana [72]

Step-by-step explanation:

Solve 2x^2-11x+14=0 give me 2 ways to get the 2 values of x?

The first way is to make factors by middle-term splitting. You get (x-2)(2x-7)=0, so the solution is x=2,7/2.

The other way is to use the formula for the solution of roots. Roots= (-b+ rootD)/2a, (-b-rootD)/2a, if the equation is of the form ax^2+bx+c=0. Here D is the discriminant. D=b^2-4ac

How do I solve 12x^2+11x+2=0?

Solve for x. 5x3−2x2−47x−14=0?

How can I prove that x^2+2x+2=0?

Is the sequence X^2 + 2x -2 =0?

What are the steps to solve 2^(2x)-3(2^x) +2=0?

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First method :

By breaking 'b' factor of x into two parts using factors of 'ac' ( a *c)

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take out common factors 2x(x-2) -7 (x-2) = 0

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Second method:

make equation in the form of (x+h)^ 2 - k = 0 ; then x+h = +sqrt(k) and -sqrt(k)

which will give x as -h+sqrt(k) , +sqrt(k)

2x^2 - 11x +14 =0 wil become x^2 -11/2 x + 7 = 0

which is (x-11/4)^2 + 7- 121/16 =0

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