In an absolute value inequality, the absolute value expression must be isolated on one side first. We can do this by subtracting 21 from both sides:

\(\displaystyle |-6x+14| + 21 < 78\)

\(\displaystyle |-6x+14| + 21-21 < 78 -21\)

\(\displaystyle |-6x+14| < 57\)

This can be rewritten as the three-part inequality

\(\displaystyle -57 < -6x+14 < 57\)

Subtract 14 from all three expressions:

\(\displaystyle -57 - 14 < -6x+14 - 14 < 57 - 14\)

\(\displaystyle -71 < -6x < 43\)

Divide all three expressions by \(\displaystyle -6\), reversing the inequality symbols since you are dividing by a negative number:

\(\displaystyle -71\div (-6) >-6x \div (-6)> 43 \div (-6)\)

\(\displaystyle 11 \frac{5}{6} >x > -7 \frac{1}{6}\)

In interval notation, this is \(\displaystyle \left ( -7 \frac{1}{6}, 11 \frac{5}{6} \right )\).

\(\displaystyle 42 - |-2x+19| < 14\)

In an absolute value inequality, the absolute value expression must be isolated on one side first. We can di this by first subtracting 42 from both sides:

\(\displaystyle 42 - |-2x+19| - 42< 14 - 42\)

\(\displaystyle - |-2x+19| < -28\)

Divide by \(\displaystyle -1\), reversing the direction of the inequality symbol since we are dividing by a negative number:

\(\displaystyle \frac{- |-2x+19| }{-1}> \frac{-28}{-1}\)

\(\displaystyle |-2x+19| >28\)

This inequality can be rewritten as the compound inequality

\(\displaystyle -2x+19 < -28\) or \(\displaystyle -2x+19 >28\)

Solve each simple inequality separately. 

\(\displaystyle -2x+19 < -28\)

Subtract 19 from both sides:

\(\displaystyle -2x+19 - 19 < -28 - 19\)

\(\displaystyle -2x < -47\)

Divide by \(\displaystyle -2\), remembering to reverse the symbol:

\(\displaystyle \frac{-2x}{-2} > \frac{-47}{-2}\)

\(\displaystyle x >23.5\)

In interval notation, this is \(\displaystyle (23.5, \infty)\).

Carry out the same steps on the other simple inequality:

\(\displaystyle -2x+19 >28\)

\(\displaystyle -2x+19 -19 >28 - 19\)

\(\displaystyle -2x >9\)

\(\displaystyle \frac{-2x}{-2} < \frac{9}{-2}\)

\(\displaystyle x < -4.5\)

In interval notation, this is \(\displaystyle (-\infty, -4.5)\).

Since the two simple inequalities are connected by an "or", their individual solution sets are connected by a union; the solution set is

\(\displaystyle (-\infty, -4.5) \cup (23.5, \infty)\).

In general, an equation of depreciation looks like the following.

\(\displaystyle y=a(1-r)^t\), where \(\displaystyle a\) is the starting amount, \(\displaystyle r\) is the common ratio, and \(\displaystyle t\) is time.

For us \(\displaystyle a=1200\), \(\displaystyle r=\frac{1}{3}\), and \(\displaystyle t=t\).

\(\displaystyle y=1200\left(1-\frac{1}{3}\right)^t\)

\(\displaystyle y=1200\left(\frac{2}{3}\right)^t\)

To answer this question, we should create a table of values and compare. For calculating the difference between the y value and the y value of the line of best, take the absolute value since it won't make since if it is negative.

\(\displaystyle \begin{tabular}{cccc} x& y & line of best fit & difference between y and line of best fit\\ \hline 1 & 1 & 2.2053& 2.2053\\ 1 & 2& 2.2053 & 0.2053\\ 1 & 3 & 2.2053 & 0.7947\\ 2& 1 & 2.5074 & 1.5074\\ 4 & 3 & 3.1116 & 0.1116\\ 4 & 5 & 3.1116 & 1.8884\\ 4 & 6 & 3.1116 & 2.8884\\ 6 & 2 & 3.7158 & 1.7158\\ 6 & 3 & 3.7158 & 0.7158 \end{tabular}\)

From our table of values, we can see that the point furthest from the line of best fit is \(\displaystyle (4,6)\).

First step is to set it equal to zero

\(\displaystyle f(x)=x^3-3x\)

\(\displaystyle x^3-3x=0\)

Now factor out an \(\displaystyle x\)

\(\displaystyle x(x^2-3)=0\)

Set up the equations, and solve for \(\displaystyle x\).

\(\displaystyle x=0\)

\(\displaystyle x^2-3=0\)

\(\displaystyle x^2=3\)

\(\displaystyle x=\pm\sqrt3\)

The correct answer is (x + 4)(x – 4) 

We neen to factor x² – 16 to solve. We know that each parenthesis will contain an x to make the x². We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x² + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer. 

First, let's factor x³ – y³ using the formula for difference of cubes.

x³ – y³ = (x – y)(x² + xy + y²)

We are told that x² + xy + y² = 6. Thus, we can substitute 6 into the above equation and solve for x – y.

(x - y)(6) = 30.

Divide both sides by 6.

x – y = 5.

The original questions asks us to find x² – 2xy + y². Notice that if we factor x² – 2xy + y² using the formula for perfect squares, we obtain the following:

x² – 2xy + y² = (x – y)².

Since we know that (x – y) = 5, (x – y)² must equal 5², or 25.

Thus, x² – 2xy + y² = 25.

The answer is 25.

This can be solved by substitution and factoring.

x² – y = 34 can be written as y = x² – 34 and substituted into the other equation: x – y = 4 which leads to x – x² + 34 = 4 which can be written as x² – x – 30 = 0.

x² – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.

When a = 9, then x² + 2ax + 81 = 0 becomes 

x² + 18x + 81 = 0.  

This equation can be factored as (x + 9)² = 0.

Therefore when a = 9, x = –9.

In general, if a function has a root at x = r, then (x – r) must be a factor of f(x). In this problem, we are told that f(x) has roots at –1, 0 and 2. This means that the following are all factors of f(x):

(x – (–1)) = x + 1

(x – 0) = x

and (x – 2).

This means that we must look for an equation for f(x) that has the factors (x + 1), x, and (x – 2).

We can immediately eliminate the function f(x) = x² + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = x² – x – 2.

Let's look at the function f(x) = x³ – x² + 2x. When we factor this, we are left with x(x² – x + 2). We cannot factor this polynomial any further. Thus, x + 1 and x – 2 are not factors of this function, so it can't be the answer.

Next, let's examine f(x) = x⁴ + x³ – 2x² .

We can factor out x².

x² (x² + x – 2)

When we factor x² + x – 2, we will get (x + 2)(x – 1). These factors are not the same as x – 2 and x + 1. 

The only function with the right factors is f(x) = x³ – x² – 2x.

When we factor out an x, we get (x² – x – 2), which then factors into (x – 2)(x + 1). Thus, this function has all of the factors we need.

The answer is f(x) = x³ – x² – 2x.