Use the mirror/lens equation:

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

Where:

$\displaystyle p$ is the object distance from the mirror, which is taken to be negative

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

$\displaystyle f$ is the focal length of the mirror, which for concave mirrors is taken to be positive

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

Plug in values and solve for the focal point:

$\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}$

$\displaystyle f=18.5cm$

Use the mirror/lens equation:

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

Where:

$\displaystyle p$ is the object distance from the mirror, which is taken to be negative

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

$\displaystyle f$ is the focal length of the mirror, which for concave mirrors is taken to be positive

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

Plug in values and solve for the image distance:

$\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}$

$\displaystyle i=\frac{1}{-\frac{2}{37}+\frac{1}{10}}$

$\displaystyle i=6.5cm$

Use the relationship between focal length and radius:

$\displaystyle f=\frac{r}{2}$

Plug in values.

$\displaystyle f=\frac{37.5cm}{2}$

$\displaystyle f=18.75cm$

Based on the properties of a concave mirror, objects inside the focal length of the mirror will generate virtual images.

Use the mirror/lens equation:

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

Where:

$\displaystyle p$ is the object distance from the mirror, which is taken to be negative number

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

$\displaystyle f$ is the focal length of the mirror, which for convex mirrors is taken to be negative number

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

Plug in values:

$\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{45}$

Solve for $\displaystyle i$:

$\displaystyle i=\frac{1}{-\frac{2}{45}+\frac{1}{10}}$

$\displaystyle i=18cm$

Use the equation for magnification and solve for magnification, $\displaystyle M$:

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

$\displaystyle M=-\frac{18}{-10}$

$\displaystyle M=1.8$

Use the mirror/lens equation:

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

Where:

$\displaystyle p$ is the object distance from the mirror, which is taken to be negative

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

$\displaystyle f$ is the focal length of the mirror, which for concave mirrors is taken to be positive

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

Plug in values:

$\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}$

Solve for the image distance:

$\displaystyle i=\frac{1}{-\frac{2}{37}+\frac{1}{10}}$

$\displaystyle i=6.5cm$

Use the magnification formula:

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

Where

$\displaystyle M$ is magnification

$\displaystyle h_i$ is image height

$\displaystyle h_p$ is object height

Plug in values and solve for the image height:

$\displaystyle \frac{h_i}{5}=-\frac{6.5}{-10}$

$\displaystyle h_i=3.25cm$

Parallel rays that reflect off a concave mirror always pass through the focal point of the mirror. 

In this question, we're told that an object is positioned outside of the curvature radius of a concave mirror. We're asked to identify how the resulting image will appear.

One way to approach this problem is to draw a ray diagram. In such a diagram, a straight line is first drawn from the top of the object horizontally to the mirror. From there, the line is drawn to pass through the mirror's focal point.

A second line is then drawn from the top of the object and straight through the focal point to the mirror. Then, the line is drawn horizontally away from the mirror.

The point at which these two lines intersect will represent the top of the image. In the diagram shown below, we can see that the image will appear in front of the mirror. Hence, it is a real image. Furthermore, the image appears upside down, meaning that it has an inverted orientation. So all together, the image will be real and inverted, which makes this the correct answer.