+2446 It does not matter if the parallelogram is a rhombus; it makes no difference.
A property of a parallelogram is that opposite sides are parallel. This means that \(\overline{JK}\parallel\overline{ML}\). By the alternate interior angles theorem, \(\angle MLJ\cong\angle LJK\). This indicates that both angles are also of equal measure. Therefore, \(m\angle MLK=m\angle LJK=25^{\circ}\)
By the triangle sum theorem, the sum of all the angles in a triangle is equal to 180 degrees. Using this rule, we can solve for the measure of the remaining angle.
| \(m\angle KLJ+m\angle LJK+m\angle JKL=180\) | Substitute the known values in for the angles. |
| \(m\angle KLJ+25+130=180\) | Now, solve for the only unknown. |
| \(m\angle KLJ+155=180\) | |
| \(m\angle KLJ=25^{\circ}\) | |
Yes, you have to do basic addition and subtraction to get the answer to this problem. And as helperid1839321 mentioned, because the diagonals of the parallelogram bisect a pair of opposite angles, this figure is a rhombus.
+2446 The argument within the square root must be evaluated before taking the square root of something. Therefore, simplify the fraction first and then take the square root.
| \(\sqrt{\frac{80+1}{68-4}}\) | Simplify the numerator and denominator before proceeding. |
| \(\sqrt{\frac{81}{64}}\) | A square root rule worth knowing is that the square root of a fraction is equivalent to the square root of the numerator divided by the square root of the denominator. In other words, \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\). It is like "distributing" the square root, if you want to think of it that way. |
| \(\sqrt{\frac{81}{64}}=\frac{\sqrt{81}}{\sqrt{64}}\) | Now, simplify the numerator and the denominator. |
| \(\frac{\sqrt{81}}{\sqrt{64}}=\frac{9}{8}=1.125\) | This is your simplified answer. |
+2446 It appears as if this post was edited since Cphill's last post.
| \(2x-\frac{x}{8}+5=35\) | Subtract 5 from both sides. |
| \(2x-\frac{x}{8}=30\) | Multiply by 8 on both sides to eliminate the pesky fraction. |
| \(16x-x=240\) | Combine like terms. |
| \(15x=240\) | Divide by 15 to isolate x. |
| \(x=16\) | |
+2446 \(m=-\frac{3}{4}\) is one solution. Omi67 and heureka both assumed that m must be a constant, but nothing specifies this restriction. For example, m could be a linear term or a quadratic term.
The discriminant of a quadratic equation is the part of the quadratic formula underneath the square root symbol (which is known as the radicand). By definition, the discriminant provides information about the roots of a given equation.
\(x = {-b \pm \sqrt{\textcolor{red}{b^2-4ac}}\over 2a}\)
For quadratics, there are 3 possible cases, and the different results convey something different.
1) If \(b^2-4ac<0\), then there are no real roots.
2) If \(b^2-4ac=0\), then there is one and only one real root.
3) If \(b^2-4ac>0\), then there are two real and distinct roots.
This information will be utilized for the following problem. The second case is relevant here.
First, assume that m is a constant. Before continuing, one should identify the values of a,b, and c.
\(a=1\hspace{5mm}b=1\hspace{5mm}c=m+1\)
| \(b^2-4ac=0\) | Substitute in the known values for the variables. |
| \(1^2-4(1)(m+1)=0\) | Simplify the left hand side. |
| \(1-4(m+1)=0\) | Subtract 1 from both sides. |
| \(-4(m+1)=-1\) | Divide by -4 on both sides. |
| \(m+1=\frac{1}{4}\) | Subtract 1 from both sides to isolate m. |
| \(m=-\frac{3}{4}\) | |
Of course, the other users also got the same answer. Now, one must consider the 2nd case:
\(a=1\hspace{5mm}b=m+1\hspace{5mm}c=1\)
| \((m+1)^2-4*1*1=0\) | Simplify the left hand side by doing the multiplication. | ||
| \((\textcolor{green}{m+1})^2-\textcolor{blue}{4}=0\) | Notice that the left hand side is a difference of two squares; factor accordingly. | ||
| \((\textcolor{green}{m+1}+\textcolor{blue}{2})(\textcolor{green}{m+1}-\textcolor{blue}{2})=0\) | Simplify both the factors. | ||
| \((m+3)(m-1)=0\) | Set both factors equal to 0 and then solve each factor separately. | ||
| Add the additive inverse of the constant in both equations to both sides. | ||
| |||
One must be careful, though, because one solved for the b, the coefficient of the x-term (linear term). This means that \(m=-3x\hspace{1mm}\text{and}\hspace{1mm}m=x\).
Time to consider the 3rd and final case.
\(a=m+1\hspace{1cm}b=1\hspace{1cm}c=1\)
| \(1^2-4(m+1)(1)=0\) | Wait a second! This looks exactly the same as our first case. One already knows, without solving, that m=-3/4 |
In this case m = coefficient of the x^2-term (quadratic term), so \(m=\frac{1}{4}x^2\)
Therefore, all the solutions are \(m=-\frac{3}{4}x^2,-3x,x,-\frac{3}{4}\)
.+2446 There is a formula that relates the number of sides of a regular polygon to its interior angle. It is the following:
\(\frac{180(n-2)}{n}\)
This formula will tell you the interior angle measure, if given the number of sides. However, we have the interior angle measure! So, just solve for n, the number of sides.
| \(\frac{180(n-2)}{n}=170\) | Multiply by n on both sides to get rid of the pesky fraction. |
| \(180(n-2)=170n\) | We can divide both sides by 10 to keep the numbers relatively small. |
| \(18(n-2)=17n\) | Distribute inside of the paretheses. |
| \(18n-36=17n\) | Subtract 18n from both sides. |
| \(-36=-n\) | Divide by -1 on both sides to isolate n. |
| \(n=36\) | |
Therefore, this polygon has 36 sides.
+2446 I am ??!!??!?!?!?!?!!? Well, thank you!
I wish I could explain how I got that answer.
I will give my best attempt, I guess.
| Column 1 | Column 2 | Column 3 | Column 4 | Column 5 | |
| Row 1 | L | L | |||
| Row 2 | L | O | O | L | |
| Row 3 | L | O | O | O | L |
| Row 4 | C | C | C | C | C |
This is just one side of the table. I have color coded the O's such that if you land on an O as your second-to-last letter, then you will have a certain number of possibilities.
A red O means there are 3 possibilities.
A Blue O means there are 2 possibilities.
Hopefully, you can see why this is the case.
Let's start with the easiest: The "C" in the corner. Before I start, I will use my own style of Cartesian coordinates where it represents the intersection of a column and a row. For example, \((C1,R1)\) is (Column 1, Row 1), for short. In this case, this letter happens to be L.
There is only one case to consider now for the corner "C." It is the following:
1) \((C5,R4)\Rightarrow(C4,R3)\Rightarrow(C4,R2)\)
Look at that! I have landed on a red O, which signifies 3 possibilities. This means that the corner "C" has 3 possibilities. There are 4 instances of the corner "C," so 4*3=12
Time to consider the next case: The "C" adjacent to the corner "C"
Now, let's consider how many cases there are for the sequence \((C4,R4)\Rightarrow(C3,R3)\). Well, the number of possibilities is equal to the sum of the number of possibilities its neighbors are immediately adjacent to.
\((C3,R3)\) is adjacent to 2 red and 2 blue O's. Because there are 3 possibilities for a red one and 2 possibilities for a blue one, the number of possibilities is \(3+3+2+2=10\). However, this is only one sequence. Let's consider the next sequence of \((C4,R4)\Rightarrow(C4,R3)\Rightarrow(C4,R2)\). Oh look! This is a red "O," which has 3 possibilities, so let's add the number of possibilities together.
\(10+3=13\)
There are 4 of these in the diagram, so \(13*4=52\)
Now, let's consider the center "C." Well, we can use the same logic as before to know that \((C3,R4)\Rightarrow(C3,R3)\) has 10 possibilities. We know that there are 2 avenues to red O's, which equals 6 additional paths.
In total, that equates to \(10+6=16\) ways. There are two instances of there, so \(2*16=32\)
The last step is to add the numbers together. \(12+52+32=96\) ways.
