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2 years ago

Solution to 3/x + 2/(x+1) = 3/5x?

Mathematics

1 answer:

s2008m [1.1K]2 years ago

8 0

9514 1404 393

Answer:

x = -6/11

Step-by-step explanation:

We assume you want the solution to ...

3/x +2/(x+1) = 3/(5x)

Subtracting the term on the right, we get the form f(x) = 0.

$\displaystyle\frac{3}{x}+\frac{2}{x+1}-\frac{3}{5x}=0\\\\\frac{15-3}{5x}+\frac{2}{x+1}=0\\\\\frac{12(x+1)+10x}{5x(x+1)}=0\\\\22x+12=0\quad\text{numerator must be zero}\\\\x+\frac{6}{11}=0\quad\text{divide by 22, simplify}\\\\\boxed{x=-\frac{6}{11}}$

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If you really intend 3/x +2/(x+1) = (3/5)x as you have written, the solution is 3 irrational values for x. I find a graphing calculator a convenient way to find them.

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Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

7 0

1 year ago

Answer choices 216 cm 247 cm 297 cm 337 cm

Natalija [7]

297 cm is the total surface area of the square pyramid

7 0

3 years ago

Mark all the statements that are true.

Artyom0805 [142]

Answer:

Option A

Option C

Step-by-step explanation:

A relationship is defined as a function if and only if each element of the "domain" set is assigned only one element of the "range" set. That is, there is only one output value y assigned to each input value x.

The relation x = 3 is not a function because there are infinite output values y, assigned to the same input element x.

(3, 2), (3, 5) (3, 9) (3,10000)

Then the option A is true

Option C is also true, the equation of the line shown is

$x = 3$