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Vesna [10]
3 years ago
12

1D. The least prime that is the sum of four different prime numbers is 17 because 17 = 2 + 3 + 5 + 7. What is the least prime gr

eater than 17 that is the sum of four different prime numbers?
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

The least prime greater than 17 that is the sum of four different prime numbers is 23

The least prime that is the sum of four different prime numbers is 17

That is, 17 = 2 + 3 + 5 + 7

Prime numbers greater than 17 are 23, 29, 31. 37, 41,........

Prime numbers less than 17 are 2, 3, 5, 7, 11, 13

How can we combine the prime numbers 2, 3, 5, 7, 11, and 13 to give the least prime number greater than 17

The only possible combination is:

2  +  3  +  7  +  11   =  23

Therefore, the least prime greater than 17 that is the sum of four different prime numbers is 23

Learn more here: brainly.com/question/25664167

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Answer:

<em>Problem with the subtraction is </em><em>950 - 863 = 87.</em>

Step-by-step explanation:

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Integration using part formula<br> <img src="https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5Enlogx%7D%20%5C%2C%20dx" id="TexFor
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Integration of I= \int {x^{n}logx \, dx=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

Step-by-step explanation:

Given integral is I= \int {x^{n}logx \, dx

Take logx=t

x=e^{t}

x^{n}=e^{nt}

\frac{1}{x} dx=dt

dx=xdt

dx=e^{t}dt

I= \int (e^{nt})(t)(e^{t})\, dt

I= \int (e^{(n+1)t})(t)\, dt

Using integration by part,

I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]

Writing in terms of x

I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]

I=[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]

I=[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]

I=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

Thus,

Integration of I= \int {x^{n}logx \, dx=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

3 0
3 years ago
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