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3 years ago

Help thanks your smart !

Mathematics

1 answer:

Vsevolod [243]3 years ago

3 0

Answer:

Step-by-step explanation:

didnt mean to click add answer

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Gift shop buys cards at $3.50 each. mark them up 80%. What do they sell them for?

pantera1 [17]

Oh what's 80 percent of 3.50 well think of 100 percent is 3.50 and then keep on subtracting until you have 80 percent of what you have

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3 years ago

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This question is based on the following circle graph. The graph represents the results of a survey in which 400 people were aske

shtirl [24]

Answer:

C) 11:12

Step-by-step explanation:

22:24 which reduced to 11:12

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3 years ago

Original Price: ? Percent of Discount: 25% Sale Price: 40$

Burka [1]

X - 0.25x = 40

Original price = $53.33

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3 years ago

Help please anyone I really need it.

alekssr [168]

Answer:

[See Below]

Step-by-step explanation:

$Hey~There!$

______________________

➜ To solve this, we have to subtract:

$100-13=87$

➜ To make sure we are correct, we add:

$87+13=100$

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So your answer would be $\bold~x=87$ .

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$Hope~this~helps~Mate!\\-Your~Friendly~Answerer,~Shane$ ヅ

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4 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