Mathematics, 23.03.2021 05:10, pearlkissp1bzl8

.necesito resolver estas operaciones con factorial ayuda por favor - A= (4ǃ+2ǃ+3ǃ)/(3ǃ+2ǃ)

2. (6ǃ-4ǃ)/2ǃ

2ǃx3ǃx4ǃ/(5ǃ+4ǃ)

R = 10ǃx8ǃ/(6ǃ+5ǃ)

((x+5)ǃ(x+3)ǃ)/((x+4)ǃ+(x+3)ǃ)=6ǃ

Answers: 2

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Answers

answered: georgeena8

A. 4

B. 348

C. 2

D. 174182400

E. x = 2

Step-by-step explanation:

Given - A= (4ǃ+2ǃ+3ǃ)/(3ǃ+2ǃ)

B = (6ǃ-4ǃ)/2ǃ

C = 2ǃx3ǃx4ǃ/(5ǃ+4ǃ)

D = 10ǃx8ǃ/(6ǃ+5ǃ)

E - ((x+5)ǃ(x+3)ǃ)/((x+4)ǃ+(x+3)ǃ)=6ǃ

Proof -

(4! + 2! + 3! )/ (3! + 2!) = [4×3×2! + 2! + 3×2!] / [3×2! + 2!]

= 2! [ 4×3 + 1 + 3] / 2! [3 + 1]

= [12 + 1 + 3] / [4]

= [16] / [4] = 4

⇒(4! + 2! + 3! )/ (3! + 2!) = 4

(6ǃ-4ǃ)/2ǃ = [6×5×4! - 4!] / 2!

= 4! [6×5 - 1] / 2!

= 4×3×2! [ 30 - 1] / 2!

= 12 [29] = 348

⇒(6ǃ-4ǃ)/2ǃ = 348

2ǃx3ǃx4ǃ/(5ǃ+4ǃ) = 2!×3!×4! / [ 5×4! + 4!]

= 2!×3!×4! / 4! [ 5 + 1]

= (2×1) × (3×2×1) / 6

= 12 / 6 = 2

⇒2ǃx3ǃx4ǃ/(5ǃ+4ǃ) = 2

10ǃx8ǃ/(6ǃ+5ǃ) = 10!×8! / [ 6×5! + 5!]

= (10×9×8×7×6×5×4×3×2×1) ×(8×7×6×5!) / 5! [6 + 1]

= (10×9×8×7×6×5×4×3×2×1) ×(8×7×6) / [7]

= (10×9×8×7×6×5×4×3×2×1) ×(8×6)

= 174182400

⇒10ǃx8ǃ/(6ǃ+5ǃ) = 174182400

((x+5)ǃ(x+3)ǃ)/((x+4)ǃ+(x+3)ǃ)=6ǃ

[(x+5)(x+4)(x+3)! (x+3)!] / [ (x+4)(x+3)! + (x+3)! ] = 6!

[(x+5)(x+4)(x+3)! (x+3)!] / (x+3)! [ (x+4)+ 1] = 6!

[(x+5)(x+4)(x+3)! / ( [ x+ 5 ] = 6!

(x+4)(x+3)! = 6!

(x+4)! = 6!

x+ 4 = 6

x = 6 - 4

x = 2

Note :

x! = x(x-1)(x-2)(4)(3)(2)(1)

answered: Guest

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.

answered: Guest

c > -10.55

step-by-step explanation:

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.necesito resolver estas operaciones con factorial ayuda por favor - A= (4ǃ+2ǃ+3ǃ)/(3ǃ+2ǃ)