TheXSquaredFactor

Expand the binomials on the left-hand side of the equation and simplify:
 

Think about how you would approach this problem if you had to simplify \((3x+1)^2-2(3x+1)\) completely. You should notice during simplifying that your steps show extreme parallelism to the options. 

This looks good to me!

I think you made a mistake, HelpPls, but I am glad that you are attempting this. 

I noticed this error in your work:
 

\(\left(4\triangle\frac{3}{8}\right)=\frac{\sqrt{4*\frac{3}{8}}}{\left|4-\frac{3}{8}\right|}\Rightarrow\frac{\frac{3}{2}}{\frac{32}{8}-\frac{3}{8}}\)

The square root operator suddenly disappeared. Otherwise, though, I think you did well with evaluating this. 

A perpendicular bisector is a line that intersects a segment at a right angle and at that segment's midpoint. In order to generate a unique equation of any line, you need to know at least one point and the slope; without these, it is impossible to generate a specific equation.

The definition I provided should give a clue as to how to find one point of the perpendicular bisector. Since we know that perpendicular bisectors intersect a segment at its midpoint, we can find the midpoint of \(\overline{JK}\) :

Finding the slope requires a slightly indirect approach. We can determine the slope of the given segment and adjust with the knowledge of perpendicular lines. I need not use the slope formula here because the y-coordinates of both points are the same, which means that the segment that connects J and K is horizontal. All horizontal lines have a slope of 0. The line of the perpendicular bisector, therefore, is vertical because a vertical line is perpendicular to a horizontal line. A vertical line has an undefined slope. Look at that! We have determined the slope of the perpendicular bisector. 

Vertical lines are written in the form of x=c, where c is a constant. We know, from the information gathered from solving for the midpoint, that \(x=-1\) is the equation of the perpendicular bisector.

I may not be a lexicographer, but I do know what affixes are. An affix is a word element attached to either the beginning or ending of the root of a word with the purpose of modifying the original word's meaning. Here are a few examples of an affix. 

Look at how many words, for example, build off of the root -clos-. This root, at its core, means to "shut." Here are a few I could think of:
 

Notice how combinations of letters added to the word change the meaning of the word while still preserving the original meaning. Let's try an easier root, -spec-, which means "to see." This one has so many examples that I will not even attempt to list them all. I tried listing ones with unique affixes and lengths:

All of these words (and many others I omitted) are some variation of "to see." 

I do not think this extreme diversity of words would be possible without affixes. Imagine if there was a completely different word for every variation of -spec- or -clos-? These are only two examples. You can attempt to investigate -aqua- or -jur- or -chrono-. 

Despite this analysis, I am not sure whether or not it would change the number of words in a language. Maybe there is some other way of organizing symbols to create words that we are not implementing in the modern day; you have to start from somewhere. Maybe other people can chime in and provide their thoughts. I suspect that this is a toss-up question and that no definite answer exists. 

A line is tangent to a circle if it intersects that circle at exactly one point. That's all we need to prove. 

Therefore, the intersection point of the line and circle is located at only \((1,-2)\). Since there is only one intersection point, the line is tangent to the circle.

I was trying to guide you in the right direction. I think you misinterpreted what I was attempting. When I said that there are three ways to orient the two congruent triangles such that they both share exactly two vertices, I had intentionally excluded some for you to figure out yourself. 

Notice how those orientations only concern two vertices in particular. Notice that there are two other pairs of vertices that we have not yet considered. Using the logic from before, if there are three ways for two vertices, this would also be true for two remaining pairs of vertices as well. 

\(3*3=9 \text{ different locations}\)

1) Variance measures how spread out a particular data set is. It does this by measuring the average distance each data point is from the true mean. Variance has a formula. It is the following:
 

\(\sigma^2=\frac{\sum_{i=1}^{N}(X_{i}-\mu)^2}{N}\)

Yes, this formula looks very complicated, but it is first important to define a few variables.

\(N = \text{sample size}\\ \mu = \text{mean of population}\\ X_i=\text{term of data}\\ \sigma^2=\text{variance}\)

What is this formula really trying to tell you to do? Well, here is an elementary breakdown:

1. Find the mean of the data. 

2. Subtract each individual element from the mean 

3. Square all results from the previous step and find the sum of the squared difference. 

4. Divide by the sample size.

Let's get to it!
 

1) Find the mean of the data. 

In order to find the mean, we must first find the sum of all the data points. 

If the sum of all the data points is 5262, then the mean is this number divided by the number of data points, 24. 

\(\mu=\frac{5262}{24}=219.25\)

2) Subtract each individual element from the mean and square that result. I will replicate the table above. 

3) Square all results from the previous step and find the sum of the squared difference. 

4) Divide by the sample size

I realize that this answer differs from Cphill's. I have reviewed my work many times, and I do not believe that I have done anything incorrect. 

2) The standard deviation, thankfully, does not require this much computation. Take the square root of the previous result, \(2056.5208\overline{3}\). 

\(\sigma=\sqrt{2056.5208\overline{3}}\approx45.3489\)

I realize that this is a lot of information to take in at once. If you have any questions, then do not hesitate to ask.

\(\hspace{4mm}49\\ *\hspace{1mm}32\\ \overline{\quad\quad}\\ \hspace{4mm}98\\ 1470\\ \overline{\quad\quad}\\ 1568\)

I multiplied these numbers as if they were integers. Notice, however, that 4.9 and 3.2 have one place after the decimal. We account for this by adding in the decimal point at the combined places of both multipliers. 

4.9 has one digit after the decimal, and 3.2 has one digit after the decimal, too, so that means that final answer has 2 places of decimal. 

\(1568\Rightarrow15.68\)