What is the domain of the function y=sqrt x?

2 answers:

8 0

Square root is not defined to for negative values. For negative values, the answer of square root is a complex number which is not in Real Domain.

So any such values, for which the expression is not defined are not in the Domian. So the domian of square root of x will be All real numbers greater than or equal to 0.

This can be expressed as:

[0 , ∞)
or
x ≥ 0

Yakvenalex [24]3 years ago

3 0

For this case we have the following function:
y = sqrt x
We must check the values of x for which the function is defined.
We have then:
x> 0
Therefore, the domain of the function is given by:
[0, inf)
Answer:
[0, inf)

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Vitek1552 [10]

Part A:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi(xl-t\tau)]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

Thus, the amplitude is b = 5.90 mm = 5.9\times10^{-3} \ m

Part B:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

Thus,

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right\\ \\ \frac{1}{\lambda} = \frac{1}{l} \\ \\ \Rightarrow\lambda= l =28.0 \ cm=\bold{2.8\times10^{-1}}$

Part C:

The general form of the equation of a transverse wave is given by:

$y(x,t)=A\cos\left[2\pi\left( \frac{x}{\lambda} - \frac{t}{T} \right)\right]$

where A is the amplitude, $\lambda$ is the wavelength, and T is the period.

Given that a certain transverse wave is described by:

$y(x,t)=bcos[2\pi\left(\frac{x}{l}-\frac{t}{tau}\right)\right]$ , where b = 5.90 mm , l = 29.0 cm , and \tau = 3.90\times10^{-2} s

The wave's frequency, f, is given by:

 $f= \frac{1}{T} = \frac{1}{\tau} = \frac{1}{3.40\times10^{-2}} =\bold{29.4 \ Hz}$

Part D:

Given that the the wavelength is $2.8\times10^{-1} \ m$ and that the wave's frequency is 29.4 Hz

The wave's speed of propagation, v, is given by:

$v=f\lambda=29.4(2.8\times10^{-1})=8.232 \ m/s$

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