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3 years ago

The isosceles triangle has a perimeter of 7.5 m.

Mathematics

2 answers:

postnew [5]3 years ago

5 0

Answer:

7.5 = 2x + 2.1

Step-by-step explanation:

The perimeter of a triangle is the sum of the sides. In an isosceles triangle, two sides are the same length. If x is the length of the two sides and y is the length of the third side, then:

P = 2x + y

Given that P = 7.5 and y = 2.1:

7.5 = 2x + 2.1

RUDIKE [14]3 years ago

3 0

Answer:

2.1 + 2x = 7.5

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

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

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