#### Answer

(a) The wave is moving in the negative x-direction.
(b) The amplitude is $A = 7.00~cm$ so the particles move a distance of $7.00~cm$ from their equilibrium positions.
(c) $f = 10.0~Hz$
(d) $\lambda = 0.333~cm$
(e) $v = 3.33~cm/s$
(f) The particle that is at $y = 7.00~cm$ and $x = 0$ when $t = 0$ will oscillate back and forth between $y = 7.00~cm$ and $y = -7.00~cm$ in SHM. Note that the x-component of the particle's motion stays as $x=0$.
(g) The wave is a transverse wave.

#### Work Step by Step

(a) In general: $y(x,t) = A~cos(k~x+\omega~t)$
$y(x,t) = (7.00~cm)~cos~[~(6.00\pi~rad/cm)~x+(20.0\pi~rad/s)~t)~]$
Since the term $(kx+\omega t)$ has a positive sign, the wave is moving in the negative x-direction.
(b) The amplitude is $A = 7.00~cm$ so the particles move a distance of $7.00~cm$ from their equilibrium positions.
(c) We can find the frequency:
$f = \frac{\omega}{2\pi} = \frac{20.0\pi~rad/s}{2\pi} = 10.0~Hz$
(d) We can find the wavelength:
$\lambda = \frac{2\pi}{k}$
$\lambda = \frac{2\pi}{6.00\pi~rad/cm}$
$\lambda = 0.333~cm$
(e) We can find the wave speed:
$v = \frac{\omega}{k}$
$v = \frac{20.0\pi~rad/s}{6.00\pi~rad/cm}$
$v = 3.33~cm/s$
(f) The particle that is at $y = 7.00~cm$ and $x = 0$ when $t = 0$ will oscillate back and forth between $y = 7.00~cm$ and $y = -7.00~cm$ in SHM. Note that the x-component of the particle's motion stays as $x=0$.
(g) The particles oscillate along the y-axis while the wave moves in the negative x-direction. Since these two motions are perpendicular, the wave is a transverse wave.