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3 years ago

Which number produces a rational number when multiplied by 0.5?

Mathematics

2 answers:

maks197457 [2]3 years ago

8 0

C.

hope this helps :)

soldi70 [24.7K]3 years ago

8 0

The right answer is c. 2.22

A rational number is a number that can be expressed as the quotient of two integers, that is, the mathematical equation that model this statement is:

$\frac{p}{q}$

where p and q are integers and q is non-zero. In the choices above, only 2.22 accomplishes with this condition because:

$2.22=\frac{111}{100}$

Therefore:

$2.22 \times 0.5=\frac{111}{50} \times 0.5 = \boxed{1.11=\frac{111}{100}}$

In conclusion, 1.11 is also a rational number

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murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$