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Mademuasel [1]
2 years ago
10

What is the remainder of the following division problem? x startlongdivisionsymbol x 1 endlongdivisionsymbol. minus x to get a r

emainder of 0 1 and a quotient of 1. x 1 0 –1
Mathematics
1 answer:
irga5000 [103]2 years ago
5 0

The remainder of the expression x/(x + 1) is a negative one. Then the correct option is D.

<h3>What is division?</h3>

The division means the separation of something into different parts, sharing of something among different people, places, etc.

The expression is given below.

\rm \rightarrow \dfrac{x}{x+1}

Add and subtract 1 in the numerator, then we have

\rm \rightarrow \dfrac{x+1-1}{x+1}\\\rm \rightarrow 1 -\dfrac{1}{x+1}

Then the remainder will be a negative one.

More about the division link is given below.

brainly.com/question/369266

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Volume(1 cube) = 1

Then divide the volume of one cube into the volume of the prism,

15 ÷ 1 = 15 cubes

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Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
Arisa [49]

Answer: S_n=5(1-\dfrac{1}{n+1}) ; 5

Step-by-step explanation:

Given series : [\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}]

Sum of series = S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}]

Consider \dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}

=\dfrac{1}{n}-\dfrac{1}{n+1}

⇒ S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}]

Put values of n= 1,2,3,4,5,.....n

⇒ S_n=5(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......-\dfrac{1}{n}+\dfrac{1}{n}-\dfrac{1}{n+1})

All terms get cancel but First and last terms left behind.

⇒ S_n=5(1-\dfrac{1}{n+1})

Formula for the nth partial sum of the series :

S_n=5(1-\dfrac{1}{n+1})

Also, \lim_{n \to \infty} S_n = 5(1-\dfrac{1}{n+1})

=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5

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