Add all flowers to get denominator
4+3+4=11
now add carnations and roses
3+4=7
so your answer is
7/11 chance
You would assume that in this figure, the number of colored sections with which are not colored with respect to a " touching " colored section, would be half of the total colored sections. However that is not the case, the sections are not alternating as they still meet at a common point. After all, it notes no two touching sections, not adjacent sections. Their is no equation to calculate this requirement with respect to the total number of sections.
Let's say that we take one triangle as the starting. This triangle will be the start of a chain of other triangles that have no two touching sections, specifically 7 triangles. If a square were to be this starting shape, there are 5 shapes that have no touching sections, 3 being a square, the other two triangles. This is presumably a lower value as a square occupies two times as much space, but it also depends on the positioning. Therefore, the least number of colored sections you can color in the sections meeting the given requirement, is 5 sections for this first figure.
Respectively the solution for this second figure is 5 sections as well.
Where’s the image? There has to be an image showing the angles to see which pairs of angles are supplementary.
ANSWER
The correct answer is 
.
<u>EXPLANATION</u>
We were given the matrix equation;
![\left[\begin{array}{cc}n-1&6\\-19&m+3\end{array}\right] +\left[\begin{array}{cc}-1&0\\16&-8\end{array}\right] =\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dn-1%266%5C%5C-19%26m%2B3%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%260%5C%5C16%26-8%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%266%5C%5C-3%2640%5Cend%7Barray%7D%5Cright%5D)
.
We must first simplify the Left Hand Side of the equation by adding corresponding entries.
![\left[\begin{array}{cc}n-1+-1&6+0\\-19+16&m+3-8\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dn-1%2B-1%266%2B0%5C%5C-19%2B16%26m%2B3-8%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%266%5C%5C-3%2640%5Cend%7Barray%7D%5Cright%5D)
.
That is;
![\left[\begin{array}{cc}n-2&6\\-3&m-5\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dn-2%266%5C%5C-3%26m-5%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%266%5C%5C-3%2640%5Cend%7Barray%7D%5Cright%5D)
.
Since the two matrices are equal, their corresponding entries are also equal. we equate corresponding entries and solve for m and n.
This implies that;
We got this equation from row one-column one entry of both matrices.
Also, the row three-column three entries of both matrices will give us the equation;
Hence the correct answer is .
The correct option is option 2
