TheXSquaredFactor

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UsernameTheXSquaredFactor
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 #2
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0

Problem #1: "how close does the line 2x-5y=4 come to the point (1,5)​"

 

When measuring distance from a point to a line, it is important that you find the perpendicular distance. I think a diagram could be useful for this problem. I have provided you with one that encompasses this exact problem:

 

 

\(\overline{BC}\) is included in the equation given, \(2x-5y=4\)

d is the perpendicular distance from the desired point, A, to the line, 

 

I think that converting the equation into slope-intercept (y=mx+b) form is the way to go. To do this, solve for y:

 

\(2x-5y=4\)Subtract the x-term to the left-hand side of the equation.
\(-5y=-2x+4\)Divide by -5.
\(y=\frac{2}{5}x-\frac{4}{5};\\ m_{\overline{CB}}=\frac{2}{5},b=-\frac{4}{5}\) 
  

 

When the equation is in this form, it gives us much more valuable information that standard form does. For example, we now know what the slope is. This can also allow us to find the slope of the perpendicular line because perpendicular lines are always opposite reciprocals of the original slope. 

 

\(m_{\overline{AB}}=\frac{2}{5}\Rightarrow -\frac{2}{5}\Rightarrow -\frac{5}{2}\)Now that we know the slope of this line, let's substitute this into the new equation for the segment.
\(y=-\frac{5}{2}x+b\)There are a few methods to determine the "b." I will just substitute the coordinate that I know that lies on this line, (1,5), and solve for b.
\(x=1,y=5;\\ 5=-\frac{5}{2}*1+b\)Solve for b.
\(5=-\frac{5}{2}+b\)Add 5/2 to solve completely. 
\(b=5+\frac{5}{2}=\frac{10}{2}+\frac{5}{2}=\frac{15}{2}\)Therefore, the final equation of the line is the following.
\(y=-\frac{5}{2}x+\frac{15}{2}\) 
  

 

Why would we want to find the equation of the line AB. Well, I now have a system of equations, so I can figure out where both lines intersect. As a refresher, here are the equation of both lines again. Both lines are already solved for y, so we can utilize the substitution method to figure out the intersection point.:

 

\(y=\frac{2}{5}x-\frac{4}{5}\\ y=-\frac{5}{2}x+\frac{15}{2}\)

 

\(\frac{2}{5}x-\frac{4}{5}=-\frac{5}{2}x+\frac{15}{2}\)When there are so many fractions in the same problem, I would suggest multiplying by the LCM of the fractions. In this case, that would be 10. 
\(10(\frac{2}{5}x-\frac{4}{5})=10(-\frac{5}{2}x+\frac{15}{2})\)Now, simplify! In the meantime, kiss those fractions goodbye!
\(2*2x-2*4=5*-5x+5*15\)This is the beauty of multiplying the entire equation by the LCM. 
\(4x-8=-25x+75\)Now, solve for x.
\(29x=83\)This system of equations is ugly, unfortunately. No nice numbers.
\(x=\frac{83}{29}\) 
  


Unfortuntately, these are not the nicest numbers. We now have to compute the y-coordinate. Substitute in the x-coordinate for one of the two equations.

 

\(y=\frac{2}{5}x-\frac{4}{5}\)x=83/29.
\(y=\frac{2}{5}*\frac{83}{29}-\frac{4}{5}\)It is possible that the best way to compute this by hand is to factor out 2/5 temporarily. 
\(y=\frac{2}{5}\left(\frac{83}{29}-2\right)\)Now, do the subtraction outside of the parentheses. 
\(\frac{83}{29}-2=\frac{83}{29}-\frac{58}{29}=\frac{25}{29}\)Plug this into the original computation. 
\(y=\frac{2}{5}*\frac{25}{29}=\frac{2}{1}*\frac{5}{29}=\frac{10}{29}\)With this, I am able to do some simplification before multipyling thankfully. 
\(\left(\frac{83}{29},\frac{10}{29}\right)\)This is the coordinate of the intersection, or point B. 
  


To find the distance, just use the distance formula! This finds the distance from point A to B, which is also the closest. 

 

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)The two coordinates we are plugging in is  \(\left(\frac{83}{29},\frac{10}{29}\right)\) and (1,5)
\(d=\sqrt{\left(\frac{83}{29}-1\right)^2+\left(\frac{10}{29}-5\right)^2}\)I would just plug this into a calculator and solve. 
\(d=\sqrt{\frac{54^2}{29^2}+\frac{135^2}{29^2}}\)Combine the fractions together. 
\(d=\sqrt{\frac{21141}{29^2}}\)Distribute the square root symbol to the numerator and denominator. 
\(d=\frac{\sqrt{21141}}{\sqrt{29^2}}\)This is really convenient because the square root of something squared is itself. We must factor the numerator, though, to ensure it has no common factors. 
\(d=\frac{\sqrt{3^6*29}}{29}\)That's a lot of factors of three! Let's bring that to the outside. 
\(d=\frac{3^3\sqrt{29}}{29}=\frac{27\sqrt{29}}{29}\approx 5.01\text{units}\)The radical form is the exact answer, and 5.01 units gives you an idea of the approximate distance. 
  

 

Problem #2: "how close does the circle with the radius 4 and center at (4,3) come to the point (10,3)?"

 

I think this question is easier to answer. The center of the circle lies on (4,3). The radius is 4 units long, so it could extend to the point (8,3), which is 2 units away from the desired point (10,3). I know this because both points lie on the same y-coordinate. 

TheXSquaredFactor May 19, 2018
 #1
avatar+1961 
0

The empirical rule (sometimes referred to as 68-95-99.7% rule) will come be handy here. This empirical rule expresses the percent of the values that are 1 to 3 standard deviations in both directions of the mean with a normal distribution:

 

Source: https://i2.wp.com/www.statistics-made-easy.com/wp-content/uploads/2016/08/empirical-rule.png?w=382&ssl=1

 

As the problem mentions, the standard deviation here is 4, and mean is 64. Therefore, one standard deviation in both directions would yield 60 and 68. The problem wants to know the percentage of values that fall in this range. As you can tell from the image above, 68% of data falls in this range. 

 

Note: The empirical rule can only be utilized in very specific circumstances. It just so happened that I could use it here. If a data value is 1.5 standard deviations away, then other methods would have to be used. 

TheXSquaredFactor May 18, 2018
 #4
avatar+1961 
+2
TheXSquaredFactor May 15, 2018
 #1
avatar+1961 
+1

An accompanying diagram is an asset geometrists use to condense large amounts of information into something more manageable and visual. I have created a reference that suits this situation perfectly. 
 

 

Notice the features of this diagram and how it relates to the actual problem:

 

  • The circle has a radius of 8 centimeters
  • There is a central angle of 90°
  • The intercepted arc, \(\widehat{BC}\), is present

Now that the important features are obvious and present, we can actually move on to the solving bit. 

 

One elementary fact about a circle is that its entire circumference can be represented with this formula:
 

\(\text{Circumference}=2\pi r\)

 

However, there is a slight issue with this formula for this particular context: The problem only asks for a certain portion of the circumference--not the whole thing. How can we remedy this issue? Well, let's just pretend for a moment that we actually knew that certain percentage, say 20%. 

 

If we knew that the problem only wanted 20% of the circumference, then we would just multiply the circumference by that arbitrary percentage, 20 in this case. The formula would be \(\text{Circumference}=0.2*2\pi r= 0.4\pi r\)

 

Ok, we are making some great progress on this problem! We still need a way of determining that percentage, though. I will introduce another well-known fact about circles: They have 360 degrees in total. We also know the central angle measure, which is 90°. This means that 90 out of the 360 degrees encompass the subtended (also referred to as intercepted) portion. This means that the "certain percentage" is determined by the central angle measure. We can generalize this and create a formula for this!

 

\(\text{Arc Length}=2\pi r*\frac{\theta}{360}\) This is what we have determined up to now. \(\theta\) is generally the Greek letter used to denote a variable in a situation similar in nature to how it is in this problem. There is only one bit of simplification possible here. 
\(r=8;\theta=90\\ \text{Arc Length}=\frac{\pi r\theta}{180}\) Substitute in the values and solve.
\(\widehat{BC}=\frac{8*90*\pi}{180}=4\pi\text{cm}\) You are finished!
   
TheXSquaredFactor May 14, 2018
 #7
avatar+1961 
+1

This post challenges my overall mentality when I answer someone's question, but it definitely shifts it for the better.

 

"Are you seriously trying to help the asker to learn?"

 

For me, the answer is patently obvious: I am undoubtedly here for the sole purpose of aiding those who may struggle in a subject where I excel. I never answer a question without providing an accompanying solution that is both comprehensive and triple verified. If I have any doubt, then I always attempt to candidly communicate any presence of uncertainty beforehand. My track record proves this. My most recent answers to date prove this again and again. 

 

For example, I think it might be important to dissect one of my answers and truly analyze the detail I put in. The horizontal lines enclose my analysis of the effectiveness of my answer compared to an unsatisfactory sample answer. Read it if you wish.

 

Sample Answer #1:

 

The sedan has a flat rate of $30.00 and a $0.50-per-mile charge, so the equation would be \(C(m)=30.00+0.5m\) . Plug in 130 for C(m) and solve for m. 

 

Sample Answer #2: 

 

Glace at the solution I provided for CarlsBolton.

 

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The difference is astounding. Let's analyze my answer and compare it to "Sample Answer #1" to see how I care.

 

In my answer, I read the question in its entirety and pointed out a somewhat common formatting error where the question does not specify which variable letter represents the total number of miles driven. I am intentionally forthright about this, and I cite the precise location of the error. I, then, state my assumption: The number of miles driven will be m since that seems to be an appropriate letter. Note how Sample Answer #1 does not acknowledge any of this in the slightest. 

 

The following paragraph could be considered unnecessary, but I generally take this structured approach when answering word problems for others. I place, in a bulleted list, the tidbits of information that I will be taking into account. By doing this, I should be converting a potentially convoluted mishmash of English paraphernalia into simple phrases that are easier to handle and grasp. While Sample Answer #1 mentions the information given, it does not provide an explanation of the correlation between the information mentioned and the generated equation. This is something my answer does with the following paragraphs. 

 

Paragraph #3 is one where I attempt to explain vocabulary (flat rate) because I realize it is key to figuring out the entire problem. I first provide the denotation, but I do not stop there. I go the extra step and explain how the vocabulary word relates to the given context of the problem. This is attention to detail. 

 

In paragraph #4, I mostly repeat what I did for the third paragraph. I explain how the rate works. I use the same structure to convey the meaning of the $0.50-per-mile rate. I have probably spent an hour generating that response (because I try to be meticulous with the writing), and I still have not even begun to answer the question.

 

In paragraph #5, I finally introduce the final equation. However, I do not do so until I perform a final consolidation of all the information I have provided the reader. Afterward, I finally reveal the equation for the particular problem.  I solve for m shortly after.

 

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If you read in between the horizontal lines, you will have seen what great lengths I had taken to provide CarlsBolton the answer. This is only one example of the 800 that I have answered thus far. I think the answer to Melody's question is so patently obvious. 

 

"Are you mostly interested in showing your own cleverness or increasing your own points?"

 

Yes, I think I have been interested in demonstrating my great abilities at mathematics. Yes, I am interested in acquiring the 2000-point milestone, which is not too far off. Maybe I should not pay attention to it because those points show how much I have let others cheat.

 

"Are you doing homework for students which is facilitating cheating and making it unlikely that they will attempt to learn anything? "

 

Yes, I am doing homework for other students. The title for CarlsBolton's question is "Need help with math homework." Yes, it is facilitating cheating. As of the writing of this post, CarlsBolton has not replied in any way to my thorough response, so I will assume that I have just made it "unlikely that [he or she] will attempt to learn anything." Unfortunately, I am part of the problem. 

 

"Personally, I do not think people who post multiple questions in close time proximity are attempting to learn anything.

This is especially when they are posted with minimal effort all in the same question post.

 I think they are just posting their homework or assignment so that they can cheat and learn nothing."

 

I wholeheartedly agree.

 

"Before you answer a question I urge you to consider WHY the question/s is asked and WHY you are providing an answer.

Do you really want to discourage people from learning and facilitate cheating?"

 

I will. I really have to contemplate what I have been doing for the past year. I have been facilitating cheating. I have been blindsided by my own generosity. That hurts.

TheXSquaredFactor May 12, 2018
 #3
avatar+1961 
+4

I recommend finding the vertex because doing this should make the process easier. This inequality is much harder than the previous one, but do not fret! It is still perfectly manageable by hand. You just need some patience. 

 

As a quick refresher, for a quadratic in the form \(ax^2+bx+c\), the coordinate of the vertex is located at \((\frac{-b}{2a},f(\frac{-b}{2a}))\).  Let's figure those coordinates right now!

 

\(-2x^2-8x-12\) This is the original expression.
\(a=-2;b=-8\\ x=\frac{-(-8)}{2*-2}=\frac{8}{-4}=-2\) This work shows the x-coordinate of the vertex. Plug in this x-coordinate to determine the y-coordinate of the vertex. 
\(x=-2\\ y=-2(-2)^2-8(-2)-12\) Let's just simplify this. The calculations should not be too difficult to do by hand.
\(y=-2(-2)^2-8(-2)-12\\ y=-2*4+16-12\\ y=-8+16-12\\ y=8-12\\ y=4\)  
\((-2,-4)\) This is the coordinate of the vertex. 
   

 

The vertex, after all, is a coordinate of the parabola, so we can pencil this coordinate onto the graph. Let's think of x-coordinates that are generally friendly to plug in. x=0 and x=1 seem like excellent candidates. When graphing any parabola, I generally strive to graph at least 5 points. Graphing at least 5 points allows me to understand the general shape of the curve.

 

Let's calculate the output when x=0:

\(x=0\\ y=-2x^2-8x-12\) Replace all instances of x with 0 to find the corresponding y-coordinate. A lot of cancellation will occur here. This will ease the process. If given the choice, always choose x=0 as a point to plug in for parabolas. It is very easy.
\(y=-2*0^2-8*0-12\\ y=-12\)  
\((0,-12)\) This is another point of the parabola. Plot this. 
   

 

Let's calculate the output when x=1:

 

\(x=1\\ y=-2x^2-8x-12\) Do the same process as before. It is not as easy to substitute in as 1, but it should be relatively easy.
\(y=-2*1^2-8*1-12\\ y=-2-8-12\\ y=-22\)  
\((1,-22)\) Graph this coordinate as well.
   

 

Remember the vertex you found earlier? This is where that calculation becomes handy. A vertical line from the vertex has a special name for quadratics: the axis of symmetry. The axis of symmetry indicates that the graph should be symmetrical about this particular line. We know that the vertex is at \((-2,-4)\). I know that \((0,-12)\) is a point, so \((-4,-12)\) is also a point on the parabola. Using the same logic for \((1,-22)\)\((-5,-22)\) is also a point. Without knowing the vertex, this shortcut would be impossible. Now just shade all the points lower than these outputs. You are done. 

TheXSquaredFactor Apr 30, 2018
 
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