Question 1209870
Let's break down each scenario:

**Key Properties:**

* **Even Function:** f(-x) = f(x)
* **Odd Function:** f(-x) = -f(x)

**1. If f(x), g(x), and h(x) are all even functions, then f(x) + g(x) + h(x) is an ___ function.**

* Let's test:
    * [f(-x) + g(-x) + h(-x)] = f(x) + g(x) + h(x)
* Therefore, f(x) + g(x) + h(x) is an **even** function.

**2. If f(x) and g(x) are odd functions and h(x) is an even function, then f(x)*g(x) + h(x) is an ___ function.**

* Let's test:
    * f(-x) * g(-x) + h(-x) = (-f(x)) * (-g(x)) + h(x) = f(x) * g(x) + h(x)
* Therefore, f(x)*g(x) + h(x) is an **even** function.

**3. If f(x) is an even function and g(x) and h(x) are odd functions, then f(x)*(g(x) + h(x)) is an ___ function.**

* Let's test:
    * f(-x) * (g(-x) + h(-x)) = f(x) * (-g(x) - h(x)) = -f(x) * (g(x) + h(x))
* Therefore, f(x) * (g(x) + h(x)) is an **odd** function.

**4. If f(x) and h(x) are odd functions and g(x) is an even function, then f(x)*g(x)*h(x) is an ___ function.**

* Let's test:
    * f(-x) * g(-x) * h(-x) = (-f(x)) * g(x) * (-h(x)) = f(x) * g(x) * h(x)
    * Now, lets test by negating all x's.
    * f(-x) * g(-x) * h(-x) = (-f(x)) * g(x) * (-h(x)) = f(x) * g(x) * h(x)
    * f(-x)g(-x)h(-x) = (-f(x)) * g(x) * (-h(x)) = f(x)g(x)h(x)
    * f(-x)g(-x)h(-x) = (-1)f(x) * g(x) * (-1)h(x) = f(x)g(x)h(x)
    * Now lets test by negating one x at a time.
    * f(-x)g(x)h(x) = (-f(x))g(x)h(x) = -f(x)g(x)h(x)
    * f(x)g(-x)h(x) = f(x)g(x)h(x)
    * f(x)g(x)h(-x) = f(x)g(x)(-h(x)) = -f(x)g(x)h(x)
    * f(-x)g(-x)h(x) = (-f(x))g(x)h(x) = -f(x)g(x)h(x)
    * f(-x)g(x)h(-x) = (-f(x))g(x)(-h(x)) = f(x)g(x)h(x)
    * f(x)g(-x)h(-x) = f(x)g(x)(-h(x)) = -f(x)g(x)h(x)
    * f(-x)g(-x)h(-x) = (-f(x))g(x)(-h(x)) = f(x)g(x)h(x)
    * The product of two odd functions is an even function, and the product of an even and an odd function is an odd function.
    * so odd * even * odd = even * odd = odd.
* Therefore, f(x) * g(x) * h(x) is an **odd** function.