Fragen 2
Antworten 592


The final four statements are explained here at


The first statement is true because corresponding sides are proportional.


Yes, it is possible. After "Y1=" command line, input \(x^2/(x\geq0)\).


Let me explain how this works, so you understand, and you don't just use rote memorization. To access the inequality symbols, press, in order, the blue second button followed by the button reading "math."


Upon doing the steps above, your interface should look like the following:



1: =

2: ≠

3: >

4: ≥

5: <

6: ≤


Use your directional arrows to navigate to the individual inequalities symbols. 


Another skill you must know how to do is storing values as variables. Use this syntax as a guideline for you.




Let's perform some extremely simple logic calculations with the calculator. 


1. Input 10→X on the homescreen command line. 

2. Input X≥0 on the homescreen command line.


What did the calculator output? That's right! The calculator outputted a 1. This is the calculator method of saying "true." Now, let's perform another experiment. It is nearly the same process as before.


1. Input -10→X on the homescreen command line. 

2. Input X≥0 on the homescreen command line.


The calculator outputted a "0" this time to signify a false statement. An interesting fact about this output is that you can also do arithmetic with this information, too. Syntax is extremely important here, so use parentheses even if you think it is completely unnecessary.


1. Input 9*(X≥0)+4


The calculator outputted 4. How did it do that, you ask? Well, let me explain.


\(9*\textcolor{red}{(x\geq0)}+4\)The calculator first evaluates inside of the parentheses so that it adheres to the order of operations.
\(\textcolor{red}{9*0}+4\)Of course, x is set to be equal to -10, so the calculator outputs a "0" because \(-10\ngeq0\). Now, the calculator continues like normal. It, of course, does multiplication first.
\(\textcolor{red}{0+4}\)Now, the calculator adds.


So, how does this help me understand what occurs for the restricted domain? Well, let's look at it. Let's pick a number that is less than 0. I'll choose -2, in this case.


\(y_1=x^2/(x\geq0)\)Since x=-2, substitute that in for every value of x.
\(y_1=(-2)^2/(-2\geq 0)\) 
\(y_1=4/0\)Uh oh! Dividing by zero causes an error, so this point is not graphed!


Now, let's try plugging in 2.


\(y_1=x^2/(x\geq0)\)Replace every instance of x with a 2.
\(y_1=2^2/(2\geq 0)\)Now, evaluate.
\(y_1=4/1\)Notice how this does not cause a problem because dividing by 1 does not actually affect the output of the function.
\(y_1=4\)The calculator would then plot that point. The process continues.


The moral here is that a false statement causes a division by zero error while a true statement leaves the original function untouched.


The way I would approach this problem is to find a formula for the sum of consecutive integers 1,2,3,4 ..., n. 


Let S = sum of consecutive whole numbers starting with 1


\(S=1+2+3+4+...+n \)


Now, let's reverse the sum. You will see where this is headed in a moment.


\(S= n+(n-1)+(n-2)+(n-3)+...+1\)


Both of the sums above are the same. Now, let's add them together.


\(S= 1+\hspace{1cm}2\hspace{2mm}+\hspace{1cm}3\hspace{2mm}+\hspace{9mm}4\hspace{2mm}+...+n\\ S=n+(n-1)+(n-2)+(n-3)+...+1\) Add these sums together.
\(2S=\underbrace{(n+1)+(n+1)+(n+1)+...+(n+1)}\\ \hspace{3cm}\text{n occurences of n+1}\) Knowing that there are n-terms present, we can simplify this slightly.
\(2S=n(n+1)\) Divide by 2 on both sides to figure out the sum of consecutive whole numbers up to n.


Now that we have generated a formula (a well-known one at that), we can proceed with this problem. I have proved this sum. 


We know that the sum must be equal to 666, so let's plug that in and solve.


\(666=\frac{n(n+1)}{2}\) Let's eliminate the fraction immediately by multiplying by 2 on both sides.
\(1332=n(n+1)\) Expand the right hand side of the equation.
\(1332=n^2+n\) Move everything to one side.
\(n^2+n-1332=0\) I will tell you beforehand that this quadratic is factorable, but it is pretty difficult to factor, so let's try using the quadratic formula instead.
\(n=\frac{-1\pm\sqrt{1^2-4(1)(-1332)}}{2(1)}\) Now, simplify
\(n=\frac{-1\pm\sqrt{5329}}{2}\) The radicand happens to be a perfect square. 
\(n=\frac{-1\pm73}{2}\) Now, solve for both values of n.
\(n=\frac{-1+73}{2}\) \(n=\frac{-1-73}{2}\)


\(n=\frac{72}{2}=36\) \(n=\frac{-74}{2}=-37\)




We reject the negative answer because we only care about whole-number solutions, and -37 is not a whole number. 


What this means is that \(666=1+2+3+4...+36\), so one must add consecutively 36 positive whole numbers, starting with 1, to get a sum of 666.


helperid1839321, I decided to delve deeper into this subject. You may like my findings!


First of all, I think I understand why this property occurs. I happen to know that \(142857\) is the first 6 digits of the decimal expansion of \(\frac{1}{7}\). At first, I thought this was insignificant, but it turns out that this fact can be used to understand this further. Look at the table below.


\(\frac{1}{7}=\overline{0.142857142857}\) Multiply both sides by 10.
\(\frac{10}{7}=1.42857\overline{142857}\) Let me rewrite 10/7 to make things clearer.
\(1+\frac{3}{7}=1.42857\overline{142857}\) Subtract one from both sides.
\(\frac{3}{7}=0.42857\overline{142857}\) WOAH! It cycles! Let's do this again. Multiply both sides by 10 again.
\(\frac{30}{7}=4.2857\overline{142857}\) Rewrite 30/7 again.
\(4+\frac{2}{7}=4.2857\overline{142857}\) Subtract 4 from both sides.
\(\frac{2}{7}=0.2857\overline{142857}\) WOAH! The first 6 digits cycle again. You can continue the pattern, if you wish!



This forced me to wonder if there is any more solutions for \(\frac{1}{p}\), where p is a whole number that creates this special property. This is because if another number p causes some repetition, then we would have found another number! YAY!


I decided to enlist some help from a computer here. This is what the computer outputted. 


7, 17, 19, 23, 29, 47


WHAT! There are more! Yes, these have the same property. Let's check them out, shall we?


\(p\) \(\frac{1}{p}\)    
7 .142857...    
17 .0588235294117647...    
19 .052631578947368421...    
23 .0434782608695652173913...    
29 .0344827586206896551724137931...    
47 .0212765957446808510638297872340425531914893617...    


It appears as if 142857 is the only number that does not start with a zero. Let's keep running this simulation! Thankfully, more numbers output! I let it run for some time, too. 


59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193...


In case you are wondering,



There are a few patterns that I see here 


1) \(p\) must be a prime number


2) The decimal expansion must have a maximum period decimal expansion of \(p-1\).


I am also making a conjecture here that I do not know whether or not is true: there are an infinite number for p that create these types of numbers!


In order to figure this one out, we will utilize some bits of manipulation.


The y-intercept of a polynomial function is determined by the constant term of each polynomial. We know, however, that the y-intercept is located at \((0,-5)\). Knowing this, we have already eliminated the last answer choice, \(y=x^4+4x^3-2x^2-5x\) because its y-intercept is located at the origin. 


The next rule I will use is something called the Descartes' rule of signs. We know, based on the given info, that at least one zero of this polynomial is negative. 


For the three remaining candidates, let's plug in the function \(f(-x)\) and count the number of sign changes. If you don't mind, I converted all the equations to function notation. 


  Candidate Function 1 Candidate Function 2 Candidate Function 3
Function \(f(x)=x^3-4x^2+2x-5\) \(f(x)=x^3+4x^2-2x-5\) \(f(x)=x^3+4x^2+2x-5\)



\(f(-x)=-x^3-4x^2-2x-5\) \(f(-x)=-x^3+4x^2+2x-5\) \(f(x)=-x^3+4x^2-2x-5\)
Number of Sign Changes in \(f(-x)\) 0 2 2


Why am I doing this, you may ask? Well, the Descartes' rule of signs states that the number of sign changes in \(f(-x)\) equals the amount of negative zeros a function has or less than by an even number. 


We can make a table to show to show the amount of negative zeros that each function can have. I'll put it in table format for your convenience.


  \(f(x)=x^3-4x^2+2x-5\) \(f(x)=x^3+4x^2-2x-5\) \(f(x)=x^3+4x^2+2x-5\)
Number of Negative Zeros 0 2 or 0 2 or 0


Wow, we have eliminated \(y=x^3-4x^2+2x-5\) from the candidates because it has no real negative roots. However, this is even more helpful than it may appear at first. Now, we will finally plug in our first point, \((-1,0)\) and see if results in a true statement. I'll choose \(x^3+4x^2+2x-5\)


\(y=x^3+4x^2+2x-5\) Plug in y=0 and x=-1. These are friendly numbers, so this should be a quick process.
\(0=(-1)^3+4(-1)^2+2*-1-5\) Evaluate this as true or false.
\(0=-1+4-2-5\) Continue simplifying.
\(0=-4\) This is an untrue statement.


This is very useful information. I now know that \((-1,0)\) does not satisfy \(y=x^3+4x^2+2x-5\), which means that this point cannot lie on this line. By process of elimination, this means that \(y=x^3+4x^2-2x-5\), the second answer choice from the top, is the correct equation.


Ok, thanks for replying. One way to check if your solution is right is to plug it in and see if the resulting equation is indeed a true one. Let's do that!




Let's plug in x=4 and x=10 and see if we get a true statement.


Check x=4

\(\sqrt{2*4-4}-4+6=0\) Let's evaluate what is inside the radical to start and determine whether or not this is a true statement. 
\(\sqrt{4}-4+6=0\) The radicand, 4, is a perfect square, so this can be simplified further.
\(2-4+6=0\) This is elementary addition and subtraction now.
\(4=0\) This is an false. statement, so this solution does not satisfy the original. 


Check x=10

\(\sqrt{2*10-4}-10+6=0\) Ok, simplify inside the radical just like before determine if true just like before.
\(\sqrt{16}-10+6=0\) Just like in the previous problem, the number underneath the square root is a perfect square, so the presence of them go away.
\(0=0\) This is a true statement, so this is a solution.


Ok, I have verified that x=10 is a valid solution, but x=4 does not satisfy this equation. I can tell you that you are wrong. The only issue here is that I cannot determine if x=4 is indeed an extraneous solution or not. I'll just solve the equation.


\(\sqrt{2x-4}-x+6=0\) Add everything but the radical expression to the right hand side of the equation.
\(\sqrt{2x-4}=x-6\) Square both sides to eliminate the radical.
\(2x-4=(x-6)^2\) Expand the right hand side of the equation.
\(2x-4=x^2-12x+36\) Move everything to one side of the equation again.
\(x^2-14x+40=0\) This is a quadratic, so there will be another solution to this equation. Factoring is easiest here.
\((x-10)(x-4)=0\) Now, set each factor equal to 0 and solve for x.
\(x-10=0 \quad x-4=0\\ \quad\quad\hspace{1mm}x=10\quad\quad\hspace{2mm} x=4\)  


Ok, you solved the equation correctly, but x=4 is an extraneous solution. However, there is no option choice that states both that x=4 is a solution and x=10 is an extraneous one.