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3 years ago

How to solve this problem (x^n+y^n) divisible by (x+y)

1 answer:

8 0

$Use \\\\a^n + b^n = (a+b)(a^{n-1}b^0 - a^{n-2}b^1 + a^{n-3}b^2 - ... - a^1b^{n-2} + a^0b^{n-1}) \\--------------------------------\\\\\dfrac{x^n+y^n}{x+y}=\dfrac{(x+y)(x^{n-1}y^0-x^{n-2}y^1+x^{n-3}y^2-...-x^1y^{n-2}+x^0y^{n-1})}{x+y}\\\\=\boxed{x^{n-1}-x^{n-2}y+x^{n-3}y^2-...-xy^{n-2}+y^{n-1}}$

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54 is 90% of what number?

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The answer is 60 surprisingly

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3 years ago

Given the geometric sequence where a1 = 3 and the common ratio is −1, what is the domain for n?

liubo4ka [24]

All integres n >= 1

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3 years ago

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1. En una biblioteca quieren comprar una computadora que

miss Akunina [59]

Answer:

678.74

that might be the question since I don't really understand what it says really

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3 years ago

Please Help I don't understand at all !!!!!!! I'll give 50 brainly points

kherson [118]

Answer:

0 People initially had the cold.

6 People will have it after 2 days.

Step-by-step explanation:

To find out how many people had it initially, plug 0 in for x (implying 0 days later.)

C= 1*3(0)

C= 1*0

C=0

Nobody had the cold initially.

To find out how many people had it after 2 days, plug 2 in for x. (implying 2 days later.)

C= 1*3(2)

C= 1*6

C=6

Therefore 6 people had the cold after 2 days.

I hope this helps!

4 0

3 years ago

Choose whether it's always, sometimes, never

Keith_Richards [23]

Answer: An integer added to an integer is an integer, this statement is always true. A polynomial subtracted from a polynomial is a polynomial, this statement is always true. A polynomial divided by a polynomial is a polynomial, this statement is sometimes true. A polynomial multiplied by a polynomial is a polynomial, this statement is always true.

Explanation:

The closure property of integer states that the addition, subtraction and multiplication is integers is always an integer.

If $a\in Z\text{ and }b\in Z$ , then a+b\in Z.

Therefore, an integer added to an integer is an integer, this statement is always true.

A polynomial is in the form of,

$p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0$

Where $a_n,a_{n-1},...,a_1,a_0$ are constant coefficient.

When we subtract the two polynomial then the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

If a polynomial divided by a polynomial then it may or may not be a polynomial.

If the degree of numerator polynomial is higher than the degree of denominator polynomial then it may be a polynomial.

For example:

$f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2$

Then $\frac{f(x)}{g(x)}=x^2+5$ , which a polynomial.

If the degree of numerator polynomial is less than the degree of denominator polynomial then it is a rational function.

For example:

$f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2$

Then $\frac{g(x)}{f(x)}=\frac{1}{x^2+5}$ , which a not a polynomial.

Therefore, a polynomial divided by a polynomial is a polynomial, this statement is sometimes true.

As we know a polynomial is in the form of,

$p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0$

Where $a_n,a_{n-1},...,a_1,a_0$ are constant coefficient.

When we multiply the two polynomial, the degree of the resultand function is addition of degree of both polyminals and the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3 0

3 years ago

Read 2 more answers