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Aug 1, 2017
 #1
avatar+2440 
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a) 

 

Let's think about this problem. Before we do, though, we must understand a few things.

 

An obtuse angle is defined by an angle wherein its measure follows the rule \(90^{\circ} . In other words, an angle that is obtuse is larger than 90 degrees and is less than 180 degrees.

 

Another thing to consider is the combined number of degrees of the interior angles of an 11-gon (known as a hendecagon). To do this, there is a formula for it. It is the following:
 

\(S=180(n-2)\)

 

Let S = sum of all interior angles in polygon

Let n = number of sides of the polygon.

 

An 11-gon has 11 sides total, so let's plug that into the formula:
 

\(S=180(n-2)\) Solve for S by plugging in the number of sides in the polygon, 11.
\(S=180(11-2)\) Simplify inside the parentheses first; 11-2=9.
\(S=180*9=1620^{\circ}\)  
   

 

I will assume that the angle I will use is 90 degrees. Divide the sum of the interior angles of degrees by 90 and see what you get. \(\frac{1620}{90}=18\). Of course, I can't have right angles as that does not fit the definition of an obtuse angle, so that means that there can only be 17 obtuse angles. 

 

But wait! An 11-gon only has 11 sides and subsequently 11 angles, which means that every angle can be obtuse!

 

B)

 

I actually found b easier than A) but that is OK.

 

There can only be, at most, 3 acute angles in a convex polygon. Only 3. This is because the exterior angle of an acute angle is greater than 90 degrees, and a 4th angle would make the exterior angle measure passed the limit of 360 degrees. Does this make sense?

 

 

 


 

 #1
avatar+2440 
0

Not all authorities agree on the standard form of a line; some believe it to be \(Ax+By=C\)while others believe it to be \(y=mx+b\). Regardless of which form you use as your standard, the answer will not be different. For this problem, I will use the form \(Ax+By=C\) because the equations are already in that form. I will, therefore, use 2 different method to solve this problem because of the likelihood of ambiguity.

 

Method 1

 

For a system, however, we have this form:

 

\(A_1x+B_1y=C_1\)

{ \(A_2x+B_2y=C_2\) 

 

Before I start solving, I need to straighten out a rule 

 

1. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\), then the system has no solution. 

2. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\), then the system has infinitely many solutions.

 

Why is this knowledge beneficial? Well, if a system has no solution or has infinitely many of them, then that means that the lines of each equation must be parallel. And by definition, if two lines are parallel, they must have the same slope. Therefore, let's look at your system:

 

{\(6x-ay=8\)

{\(2x+3y=12\)

 

Let's see what happens when we try to put the numbers over one another:
 

Does \(\frac{6}{2}=\frac{-a}{3}\neq\frac{8}{12}\)? Yes, 6/2=3, and that is definitely not equivalent to 8/12, so we know that we have the first case. To solve for a, though, we must set it equal to \(\frac{A_1}{A_2}\), which is equal to 3. Let's do that:

 

\(3=\frac{-a}{3}\) Multiply by 3 on both sides.
\(9=-a\) Divide by -1.
\(a=-9\)  
   

 

What if you want to check your answer? To check your answer, you can graph it and see if the lines ever intersect. If they do not intersect, then you have done the problem correctly.

 

You can see the graph for this method here at https://www.desmos.com/calculator/9nimq6ig4q

 

Method 2

 

For method 2, I will convert both equations to y=mx+b form and then go from there:

 

First, I will convert 6x - ay = 8:

 

\(6x-ay=8\) Subtract 6x on both sides.
\(-ay=-6x+8\) Divide by -a on both sides of the equaton.
\(y=\frac{-6x+8}{-a}\) I will break the fraction using the rule that \(\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\)
\(y=\frac{-6x}{-a}+\frac{8}{-a}\)  
\(y=\frac{6}{a}x-\frac{8}{a}\)  
   

 

Now, I will do the same to the other equation:

 

\(2x+3y=12\) Our objective is to solve for y. To accomplish this, subtract x from both sides.
\(3y=-2x+12\) Divide by 3 on both sides.
\(y=\frac{-2x+12}{3}\) Break up the fraction using the same process as above.
\(y=-\frac{2}{3}x+4\)  
   

 

Of course, the generalized equation for a line is y=mx+b where m is the slope. In order for 2 lines to have the same slope, we must set them equal to one another and solve for a:

 

\(-\frac{2}{3}=\frac{6}{a}\) Multiply by a on both sides of the equation.
\(\frac{a}{1}*\frac{-2}{3}=\frac{6}{a}*\frac{a}{1}\) Simplify both sides of the equation.
\(\frac{-2a}{3}=6\) Multiply by 3 on both sides.
\(-2a=18\) Divide by -2 on both sides.
\(a=-9\) Just like with the method above, we got the same answer
   

 

You can see the graph for this method here at https://www.desmos.com/calculator/rmv645u9ij


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