Write the number:

$\displaystyle 0.000\; 000\; 076$

Counting the number of places, move the decimal point to the right until it is after the first nonzero digit:

$\displaystyle 0 \; 000\; 000\; 07.6$

The resulting number is 7.6; the number of places the decimal point moved to the right was 8. In scientific notation, this number is $\displaystyle 7.6 \times 10^{-8}$.

Take advantage of the exponent rules.

$\displaystyle \left (7 \times 10^{15} \right )^{2}$

$\displaystyle = 7^{2} \times \left ( 10^{15} \right )^{2}$

$\displaystyle = 49 \times 10^{15 \times 2}$

$\displaystyle = 49 \times 10^{30}$

$\displaystyle = 4.9 \times 10^{1} \times 10^{30}$

$\displaystyle = 4.9 \times 10^{1+30}$

$\displaystyle = 4.9 \times 10^{31}$

Each of the answers to this question are in scientific notation. To determine the exponent on the 10, see how many places you need to move the decimal over to get from your original number to 2.79.

Because the decimal place moves 5 places, the exponent will be 5, and because the decimal moves to the RIGHT, the exponent is going to be negative.

So the answer to the quesiton is $\displaystyle 2.79\cdot 10^{-5}$

Our first value for scientific notation should be between $\displaystyle 1$ and $\displaystyle 10$. We then multiply this value by $\displaystyle 10$ raised to a power equal to the number of spaces our decimal has moved to the left.

$\displaystyle x$ cannot equal a value which would make the denominator equal to 0.  In order to figure out what that value is, we must first simplify this fraction, then set each factor of the denominator equal to 0.  As follows:

$\displaystyle \frac{x^{2}-4}{x^2-x-2}$

First, simplify by finding the original binomial multiples:

$\displaystyle \frac{(x-2)(x+2)}{(x-2)(x+1)}$

Now, set each factor of the denominator equal to 0

$\displaystyle x+1=0$

$\displaystyle x=-1$

$\displaystyle x-2=0$

$\displaystyle x=2$

If $\displaystyle x=-1$ OR $\displaystyle x=2$, the denominator will equal 0.  $\displaystyle -1$ is the only choice provided by the answers.

Note: Even though you can cancel out $\displaystyle x-2$ from the numerator and denominator, $\displaystyle x$ still cannot be equal to two. The graph would have a hole at $\displaystyle x=2$ and an aymptote at $\displaystyle x=-1$.

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:

$\displaystyle 3x-2 = 0$

$\displaystyle 3x = 2$

$\displaystyle x= \frac{2}{3}$

Therefore, the domain of $\displaystyle f$ is the set of all real numbers except $\displaystyle \frac{2}{3}$ - that is, 

$\displaystyle \left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

The domain of $\displaystyle g$ is restricted by two different denominators, neither of which can be equal to 0, so the excluded values are:

$\displaystyle x-2 = 0 \Rightarrow x = 2$

$\displaystyle x-3 = 0 \Rightarrow x =3$

The correct response is therefore $\displaystyle \left ( -\infty, 2 \right ) \cup (2,3) \cup (3, \infty)$.

The numerator, being a polynomial, does not restrict our domain. The denominator, however, does restrict it to the values for which it is not equal to 0. We set the denominator equal to 0 to find the excluded values:

$\displaystyle x^{2} - 9 = 0$

$\displaystyle (x-3) (x+3) = 0$

$\displaystyle x- 3 = 0 \Rightarrow x = 3$

$\displaystyle x+3 = 0 \Rightarrow x = -3$

The domain, in interval notation, is therefore

$\displaystyle \left ( -\infty, -3 \right ) \cup (-3,3) \cup (3, \infty)$.

The domain of $\displaystyle g$ is restricted by two things.

First, the expression within the radical in the numerator must be nonnegative. We therefore solve for $\displaystyle x$ in the inequality

$\displaystyle x + 3 \geq 0$

$\displaystyle x \geq -3$,

or, in interval notation, $\displaystyle \left [-3, \infty \right )$

Second, the expression in the denominator must be nonzero.  Therefore, we set the denominator equal to zero to determine the excluded value(s):

$\displaystyle x - 4 = 0$

$\displaystyle x = 4$

We exclude 4 from $\displaystyle \left [-3, \infty \right )$, so the correct response is 

$\displaystyle \left [-3, 4\right ) \cup \left (4, \infty \right )$

There are two things restricting the domain of $\displaystyle h$.

One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:

$\displaystyle x - 1 \geq 0$

$\displaystyle x \geq 1$,

or, in interval notation, $\displaystyle [1, \infty)$

The other is the denominator, which must be equal to 0, so set, and solve for $\displaystyle x$ in, the equation:

$\displaystyle x^{2}- 3x - 4 = 0$

$\displaystyle (x-4)(x+1) = 0$

$\displaystyle x - 4 = 0 \Rightarrow x = 4$

$\displaystyle x+ 1 = 0 \Rightarrow x = -1$

$\displaystyle -1$ is already excluded from the domain; we exclude 4, so the domain is

$\displaystyle \left [1, 4 \right ) \cup \left ( 4 ,\infty)$.