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3 years ago

An arthimetic sequence with first term 10 and common difference of 3

Mathematics

1 answer:

ipn [44]3 years ago

7 0

10, 13, 16, 19, etc. Continue adding the difference in an arithmetic sequence

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Which are the solutions of x2 = –5x + 8? SOLUTIONs ok not a single answer i need to know its opposite

AlekseyPX

Answer:

I think it is -5+root-7/2 & -5-root-7/2

Step-by-step explanation:

5 0

4 years ago

What is the median value of the data set shown on the line plot?

Mars2501 [29]

Hello!

The median is the number in the middle when the data is ordered from least to greatest

Order the numbers first

1, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 11

From here when find the medians which are two 6's which means the median is 6

Hope this helps!

4 0

3 years ago

Help me please I’m not sure about this

frutty [35]

easy, 36/6=6

6 people, right?

7 0

3 years ago

Read 2 more answers

Here is my math question: Simplify 5x^2-3x-5. I know the answer is 5x^2-13x+3, I don't know how to get to that answer because I

SIZIF [17.4K]

Here you go. I hope you understand. Ask if you don't.

5 0

3 years ago

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

liq [111]

Answer:

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