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3 years ago

Find the quotient of n(x) = 6x^5y^7 and m(x) = 2x^3y

Mathematics

1 answer:

likoan [24]3 years ago

6 0

Answer:

a. Yes

b. Yes

c. Yes

d. Degree 8

Step-by-step explanation:

a. Yes, n(x) is a polynomial of one single term (also called monomial) because it contains variables raised to positive integers.

b. Yes, m(x) is a polynomial of also one single term (also called monomial) because it contains variables raised to positive integers.

c. The quotient of n(x) / m(x) can be reduced to a polynomial of one single term as follows:

$\frac{n(x)}{m(x)} =\frac{6\,x^5\,y^7}{2\,x^3\,y} =3\,x^2\,y^6$

which as can be seen, also contains variables raised to positive integers.

d. The degree of the polynomial resultant is the addition of the powers of all variables present (x and y) which results in: 2 + 6 = 8

Therefore the degree of this polynomial is 8.

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2 years ago

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2(w – 3y + 4) + w – 5

melisa1 [442]

Answer: It should be 3w - 6y + 3

Step-by-step explanation:

-Solve:

$2(w-3y+4)+w-5$

-Use Distributive Property:

$2(w-3y+4)+w-5$

$2w-6y+8+w-5$

-Then, combine like terms:

$2w-6y+8+w-5$

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Result:

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2 years ago

We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent

Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

( X = -2 ): { X_2 = 1 , X_1 = 3 }

P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }

P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 3 / 9 ) = ( 1 / 3 )

( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

P ( X = 1 ): P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 2 ): { X_2 = 1 , X_1 = 3 }

P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

- The distribution Y = X_2,

P(Y=0) = 0

P(Y=1) = 1/3

P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

7 0

2 years ago

What are the zeros of the polynomial function f(x) = x3 + 2x2 − 24x?

Alik [6]

You may notice that all of the terms have a common factor: $x$ . So, let's take the $x$ out of all the terms. This gives us:

$x(x^2 + 2x - 24) = 0$

Using the Zero Product Property, we can say that both $x = 0$ and $x^2 + 2x - 24 = 0$ .

Let's solve both. $x = 0$ is obvious, but we still need to solve our other term:

$x^2 + 2x - 24 = 0$

Set up

$(x + 6)(x - 4) = 0$

Factor

$x = -6, 4$

Apply the Zero Product Property

Our answer is x = -6, 0, 4.

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2 years ago

Help please! Quickly!

Stells [14]

Answer:

1. x=--3

2. x=2

3. x=1/2

4. x=-1/3

Step-by-step explanation:

Hope this helps UwU

8 0

2 years ago