The Magnification Equation is as follows:

$\displaystyle M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

The negative on the last part is very important. If you don't include it (and it's easy to forget), you will get the wrong answer.

We'll use the parts without $\displaystyle M$ for our problem. We want $\displaystyle h_i$, so we just need to multiply both sides by $\displaystyle h_o$.

$\displaystyle h_i=-\frac{d_i}{d_o}\cdot h_o$

Now, we can plug in our numbers.

$\displaystyle h_i&=-\frac{d_i}{d_o} \cdot h_o$

$\displaystyle h_1=-\frac{-3\cdot 10^{-2}}{9\cdot 10^{-2}}\cdot (11\cdot 10^{-2})$

$\displaystyle h_1= 0.0367$

Therefore, the image height is $\displaystyle 3.67cm$. 

The magnification equation is 

$\displaystyle M=\frac{h_i}{h_o}$

where $\displaystyle h_i$ is the height of the image, and $\displaystyle h_o$ is the height of the object. We can plug these given values into the equation.

$\displaystyle M&=\frac{h_i}{h_o}$

$\displaystyle M=\frac{21.9}{14.7}$

$\displaystyle M= 1.49$

The magnification is a scalar value, so it's unitless. We can verify this by examining what goes into the equation. Both heights have the same units, which cancel, so that leaves us with no units.

The equation for magnification is

$\displaystyle M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

We're given the image distance and the magnification, so we'd use the second equality.

$\displaystyle M&=-\frac{d_i}{d_o}$

$\displaystyle M\cdot d_o&=-d_i$

$\displaystyle d_o&=-\frac{d_i}{M}$

Now, we can plug in our numbers.

$\displaystyle d_o&=-\frac{d_i}{M}$

$\displaystyle d_o= -\frac{-16.2}{1.79}$

$\displaystyle d_o= 9.05$

The negative in front of the equation is very important. If you forget it, the answer will be incorrect.

The distance from the mirror to the object is $\displaystyle 9.05m$. Note that the object distance is always positive.

Using the relationship:

$\displaystyle \frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}$

Where:

$\displaystyle p$ is the object distance from the mirror

$\displaystyle i$ is the image distance from the mirror

$\displaystyle f$ is the focal length of the mirror

$\displaystyle r$ is the radius of curvature of the mirror

Plugging in values:

$\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{67}$

Solving for $\displaystyle i$:

$\displaystyle i=\frac{1}{-\frac{2}{67}+\frac{1}{10}}$

$\displaystyle i=14.25cm$

Using the equation for magnification:

$\displaystyle M=\frac{h_i}{h_p}=-\frac{i}{p}$

Where:

$\displaystyle M$ is magnification

$\displaystyle M=-\frac{14.25}{-10}$

$\displaystyle M=1.425$

To determine magnification, we simply divide the object length from the image length. 

$\displaystyle M=-\frac{1.5*10^{-2}m}{10m}=-1.5*10^{-3}$

The negative sign is used to designate that the image is real.