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vlada-n [284]
3 years ago
13

I’m not sure how to do this

Mathematics
1 answer:
seraphim [82]3 years ago
7 0
\bf \cfrac{\frac{x+4}{3}+\frac{1}{x}}{5+\frac{15}{x}}\qquad \cfrac{\impliedby \frac{LCD}{3x}}{\impliedby \stackrel{LCD}{x}}\implies \cfrac{\quad \frac{x(x+4)+3(1)}{3x}\quad }{\frac{x(5)+(1)15}{x}}\implies \cfrac{\quad \frac{x^2+4x+3}{3x}\quad }{\frac{5x+15}{x}}

\bf \cfrac{x^2+4x+3}{3\underline{x}}\cdot \cfrac{\underline{x}}{5x+15}\implies \cfrac{x^2+4x+3}{3(5x+15)}\implies \cfrac{x^2+4x+3}{15x+45}
\\\\\\
\cfrac{(x+3)(x+1)}{15x+45}\implies \cfrac{\underline{(x+3)}(x+1)}{15\underline{(x+3)}}\implies \cfrac{x+1}{15}
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leonid [27]
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Read 2 more answers
Divid 15 into two parts such that the sum of their reciprocal is 3/10
emmasim [6.3K]
Let x and y be the 2 parts of 15 ==> x + y=15 (given)

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 ==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy

But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50

Now we have the sum S of the 2 parts that is S = 15 and 
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
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3 years ago
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Valentin [98]

Answer:

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7 0
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