#1**+2 **

The maximum or minimum of any quadratic is always at the input value of \(\frac{-b}{2a}\) where the standard form of a quadratic is \(ax^2+bx+c=0\) .

\(a=-4, b=16;\\ x=\frac{-b}{2a}\) | Substitute in the known values for a and b . |

\(x=\frac{-16}{2*-4}\) | Solve for x by simplifying completely. |

\(x=2\) | This is the optimal number of items sold, in thousands, in order to make maximum profit. Let's substitute that into the original function to determine the profit with that number of items sold. |

\(x=2;\\ p(2)=-4*2^2+16*2-7\) | Yet again, just simplify from here. |

\(p(2)=-4*4+32-7\) | |

\(p(2)=-16+25\) | |

\(p(2)=9\) | This is the maximum profit that this business, in thousands of dollars, makes. |

TheXSquaredFactor
Oct 14, 2018

#1**+1 **

Hi, nellycrane!

I think you will be better off if I give you an easier problem like \(x^2-4 \) . You are probably well aware that this is a difference of squares. \((x+2)(x-2)\) would be the corresponding factorization.

Now, let's say that you want to factor \(x^2-30\) . This is not possible if you restrict yourself to the rational number set. However, it can be factored as \((x+\sqrt{30})(x-\sqrt{30})\).

In general, \(a^2-b^2=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) . You can apply this knowledge to the problems at hand.

\(x^2+50\\ a=x^2,b=-50;\\ (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\\ (\sqrt{x^2}+\sqrt{-50})(\sqrt{x^2}-\sqrt{-50}) \)

You can simplify this to get the factorization amongst the complex numbers. Good luck!

TheXSquaredFactor
Sep 29, 2018

#1**+2 **

This problem is easier than it looks at first glance. If you utilize clever algebraic manipulation, this problem becomes simpler.

\(x-y=16\) | \(xy=23\) |

\((x-y)^2=16^2\) | |

\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\) | \(\boxed{2}\hspace{1mm}2xy=46\) |

Notice what I have done. I have manipulated the information I know about these real numbers, *x *and *y* , and I am manipulating it in a way that is much more convenient for this particular problem. The only thing left to do is add the equations together.

\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\\ \boxed{2}\hspace{5mm}+2xy\hspace{10mm}=46\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \hspace{18mm}x^2+y^2=302\)

TheXSquaredFactor
Sep 29, 2018

#1**+1 **

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

\((a-1)^3=a^3-3a^2+3a-1\\ (a-1)^2=a^2-2a+1\) | Let's put this together into one cohesive expression. |

\(a^3-3a^2+3a-1+a^2-2a+1\) | Combine like terms. |

\(a^3-2a^2+a\) | Factor out an a . |

\(a(a^2-2a+1)\) | The trinomial is indeed factorable and happens to factor into a perfect square. |

\(a(a-1)(a-1)\) | |

\(a(a-1)^2\) | |

TheXSquaredFactor
Sep 29, 2018

#2**0 **

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 \(\overline{JK}\) :

\(m_{\overline{JK}}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\) | This is the midpoint formula. The midpoint formula just finds the average of the two coordinates. |

\(m_{\overline{JK}}=\left(\frac{-4+2}{2},\frac{2+2}{2}\right)\) | The only thing left to do is to simplify. |

\(m_{\overline{JK}}=(-1,2)\) | Yes! We have already found one part of the perpendicular bisector, a point. |

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.

TheXSquaredFactor
Sep 28, 2018

#1**+1 **

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:

Closet |

Closely |

Closest |

Disclose |

Foreclosure |

Closeness |

Closemouthed |

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:

Retrospectively |

Specifications |

Disrespectful |

Spectrograph |

Spectacular |

Inspectors |

Suspected |

Spectrum |

Aspects |

Speck |

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.

TheXSquaredFactor
Sep 27, 2018

#1**+2 **

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

\(\boxed{1}\hspace{1mm}x=2y+5\\ \boxed{2}\hspace{1mm}x^2+y^2=5\) | We can use the method of substitution to find the point of intersection. |

\((2y+5)^2+y^2=5\) | We have to expand the square of a binomial in order to make some progress here. |

\(4y^2+20y+25+y^2=5\) | Combine the like terms and bring all terms to one side of the equation. |

\(5y^2+20y+20=0\) | Factor out the greatest common factor from every term, 5. |

\(5(y^2+4y+4)=0\) | The trinomial contained in the parentheses can be written as a square of a binomial. |

\(5(y+2)^2=0\) | Divide by 5 from both sides of the equation. |

\((y+2)^2=0\) | Take the square root of both sides. |

\(\sqrt{(y+2)^2}=\sqrt{0}\) | Write the left-hand side as the absolute value of the argument. |

\(|y+2|=0\) | Solve for y. |

\(y+2=0\) | |

\(y=-2\) | Now, solve for x. Since we are proving that this line is tangent, we should have foreseen that there would only be one possible y-value. Now, we must solve for the x-coordinate. |

\(\boxed{1}\hspace{1mm}x=2y+5; y=-2\) | Of course, you may substitute into either equation, but it is probably desirable to substitute the known y-value into this equation because it is most likely easier. |

\(x=-2*2+5\) | Simplify. |

\(x=1\) | |

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.

TheXSquaredFactor
Sep 27, 2018

#5**+2 **

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.

\(x_i\) | |

\(x_1\) | 182 |

\(x_2\) | 274 |

\(x_3\) | 207 |

\(x_4\) | 216 |

\(x_5\) | 285 |

\(x_6\) | 190 |

\(x_7\) | 138 |

\(x_8\) | 240 |

\(x_9\) | 291 |

\(x_{10}\) | 288 |

\(x_{11}\) | 176 |

\(x_{12}\) | 186 |

\(x_{13}\) | 190 |

\(x_{14}\) | 213 |

\(x_{15}\) | 236 |

\(x_{16}\) | 242 |

\(x_{17}\) | 186 |

\(x_{18}\) | 259 |

\(x_{19}\) | 164 |

\(x_{20}\) | 172 |

\(x_{21}\) | 233 |

\(x_{22}\) | 268 |

\(x_{23}\) | 152 |

\(x_{24}\) | 274 |

\(\Sigma\) | 5262 |

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.

\(x_i\) | \(\mu\) | \(x_i-\mu\) | |

\(x_1\) | 182 | -37.25 | |

\(x_2\) | 274 | 54.75 | |

\(x_3\) | 207 | -12.25 | |

\(x_4\) | 216 | -3.25 | |

\(x_5\) | 285 | 65.75 | |

\(x_6\) | 190 | -29.25 | |

\(x_7\) | 138 | -81.25 | |

\(x_8\) | 240 | 20.75 | |

\(x_9\) | 291 | 71.75 | |

\(x_{10}\) | 288 | 68.75 | |

\(x_{11}\) | 176 | -43.25 | |

\(x_{12}\) | 186 | -33.25 | |

\(x_{13}\) | 190 | -29.25 | |

\(x_{14}\) | 213 | -6.25 | |

\(x_{15}\) | 236 | 16.75 | |

\(x_{16}\) | 242 | 22.75 | |

\(x_{17}\) | 186 | -33.25 | |

\(x_{18}\) | 259 | 39.75 | |

\(x_{19}\) | 164 | -55.25 | |

\(x_{20}\) | 172 | -47.25 | |

\(x_{21}\) | 233 | 13.75 | |

\(x_{22}\) | 268 | 48.75 | |

\(x_{23}\) | 152 | -67.25 | |

\(x_{24}\) | 274 | 54.75 | |

\(\Sigma\) | 5262 | 219.25 |

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

\(x_i\) | \(\mu\) | \(x_i-\mu\) | \((x_i-\mu)^2\) | |

\(x_1\) | 182 | -37.25 | 1387.5625 | |

\(x_2\) | 274 | 54.75 | 2997.5625 | |

\(x_3\) | 207 | -12.25 | 150.0625 | |

\(x_4\) | 216 | -3.25 | 10.5625 | |

\(x_5\) | 285 | 65.75 | 4323.0625 | |

\(x_6\) | 190 | -29.25 | 855.5625 | |

\(x_7\) | 138 | -81.25 | 6601.5625 | |

\(x_8\) | 240 | 20.75 | 430.5625 | |

\(x_9\) | 291 | 71.75 | 5148.0625 | |

\(x_{10}\) | 288 | 68.75 | 4726.5625 | |

\(x_{11}\) | 176 | -43.25 | 1870.5625 | |

\(x_{12}\) | 186 | -33.25 | 1105.5625 | |

\(x_{13}\) | 190 | -29.25 | 855.5625 | |

\(x_{14}\) | 213 | -6.25 | 39.0625 | |

\(x_{15}\) | 236 | 16.75 | 280.5625 | |

\(x_{16}\) | 242 | 22.75 | 517.5625 | |

\(x_{17}\) | 186 | -33.25 | 1105.5625 | |

\(x_{18}\) | 259 | 39.75 | 1580.0625 | |

\(x_{19}\) | 164 | -55.25 | 3052.5625 | |

\(x_{20}\) | 172 | -47.25 | 2232.5625 | |

\(x_{21}\) | 233 | 13.75 | 189.0625 | |

\(x_{22}\) | 268 | 48.75 | 2376.5625 | |

\(x_{23}\) | 152 | -67.25 | 4522.5625 | |

\(x_{24}\) | 274 | 54.75 | 2997.5625 | |

\(\Sigma\) | 5262 | 219.25 | 49356.5 |

4) Divide by the sample size

\(x_i\) | \(\mu\) | \(x_i-\mu\) | \((x_i-\mu)^2\) | \(\frac{(x_i-\mu)^2}{N}\) | |

\(x_1\) | 182 | -37.25 | 1387.5625 | ||

\(x_2\) | 274 | 54.75 | 2997.5625 | ||

\(x_3\) | 207 | -12.25 | 150.0625 | ||

\(x_4\) | 216 | -3.25 | 10.5625 | ||

\(x_5\) | 285 | 65.75 | 4323.0625 | ||

\(x_6\) | 190 | -29.25 | 855.5625 | ||

\(x_7\) | 138 | -81.25 | 6601.5625 | ||

\(x_8\) | 240 | 20.75 | 430.5625 | ||

\(x_9\) | 291 | 71.75 | 5148.0625 | ||

\(x_{10}\) | 288 | 68.75 | 4726.5625 | ||

\(x_{11}\) | 176 | -43.25 | 1870.5625 | ||

\(x_{12}\) | 186 | -33.25 | 1105.5625 | ||

\(x_{13}\) | 190 | -29.25 | 855.5625 | ||

\(x_{14}\) | 213 | -6.25 | 39.0625 | ||

\(x_{15}\) | 236 | 16.75 | 280.5625 | ||

\(x_{16}\) | 242 | 22.75 | 517.5625 | ||

\(x_{17}\) | 186 | -33.25 | 1105.5625 | ||

\(x_{18}\) | 259 | 39.75 | 1580.0625 | ||

\(x_{19}\) | 164 | -55.25 | 3052.5625 | ||

\(x_{20}\) | 172 | -47.25 | 2232.5625 | ||

\(x_{21}\) | 233 | 13.75 | 189.0625 | ||

\(x_{22}\) | 268 | 48.75 | 2376.5625 | ||

\(x_{23}\) | 152 | -67.25 | 4522.5625 | ||

\(x_{24}\) | 274 | 54.75 | 2997.5625 | ||

\(\Sigma\) | 5262 | 219.25 | 0 | 49356.5 | \(2056.5208\overline{3}\) |

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.

TheXSquaredFactor
Sep 26, 2018