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3 years ago

Evaluate the factorial expression. 19! 16!(3 – 1)!

Mathematics

1 answer:

krok68 [10]3 years ago

7 0

Answer:

I believe it would be 2907

Step-by-step explanation:

starting with (3-1)! you would get 2, then going on to 16! you would get 20,922,789,888,000, then multiplying that with 2 you would get 41,845,579,776,000, then moving onto the 19! you would get 121,645,100,408,832,000, now last but not least to devide 41,845,579,776,000 by 121,645,100,408,832,000, and to that you would get 2907 hope this helped!

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USPshnik [31]

The tree in exactly A.7.5 meters

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3 years ago

The score on a trivia game is obtained by subtracting the number of incorrect answer from twice the number of correct answer. If

brilliants [131]

Answer:

The player answer correctly 30 questions

Step-by-step explanation:

First we have to develop the information that gives us as 2 equations

x = correct answer

y = incorrect answer

z = score = 40

h = total answer = 50

2x - y = z x + y = h

2x - y = 40 x + y = 50

we clear the x of one equation and replace the value of x in the other equation

x = 50 - y

2x - y = 40

2(50 - y) - y = 40

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y = -60/-3

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x = 30

3 0

3 years ago

357 miles in 6.3 hours

FromTheMoon [43]

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7 0

3 years ago

Find the sum. 13(9−6m)+14(12m−8) =

allochka39001 [22]

Answer:

5+90m

Step-by-step explanation:

117 - 78m + 168m - 112

117- 112 = 5

-78m + 168m= 90m

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7 0

3 years ago

What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i? f(x) = x2 – 2x + 2 f(x) = x3 – x

konstantin123 [22]

Answer:

$f(x)=x^{3}-3x^{2} +4x-2$

Step-by-step explanation:

we know that

The conjugate root theorem states that if the complex number a + bi is a root of a polynomial P(x) in one variable with real coefficients, then the complex conjugate a - bi is also a root of that polynomial

In this problem we have that

The polynomial has roots 1 and (1+i)

by the conjugate root theorem

(1-i) is also a root of the polynomial

therefore

The lowest degree of the polynomial is 3

$f(x)=a(x-1)(x-(1+i))(x-(1-i))$

Remember that

The leading coefficient is 1

a=1

$f(x)=(x-1)(x-(1+i))(x-(1-i))\\\\f(x)=(x-1)[x^{2} -(1-i)x-(1+i)x+(1-i^2)]\\\\f(x)=(x-1)[x^{2} -x+xi-x-xi+2]\\\\f(x)=(x-1)[x^{2} -2x+2]\\\\f(x)=x^{3}-2x^{2} +2x-x^{2} +2x-2\\\\f(x)=x^{3}-3x^{2} +4x-2$

5 0

3 years ago

Read 2 more answers