No it is not always a mixed number if you add two mixed numbers
Answer: 
Step-by-step explanation:
We know that the total number of outcomes for fair dice {1,2,3,4,5,6} = 6
Given : Favorable outcome = 5
i.e. Number of favorable outcomes =1
We know that the formula to find the probability for each event is given by :-

Then, the probability that the six faces cube will land with the number 5 face up will be :_

Hence, the required probability = 
The scale factor of 4 was applied to the first triangle. 6 was divided by 4 to get 1.5
Answer:
- 1
Step-by-step explanation:
8m + 5n
8(3) + 5(- 5)
24 + (- 25)
24 - 25
- 1
![\begin{array}{rrrrr} 10x&-&18y&=&2\\ -5x&+&9y&=&-1 \end{array}~\hfill \implies ~\hfill \stackrel{\textit{second equation }\times 2}{ \begin{array}{rrrrr} 10x&-&18y&=&2\\ 2(-5x&+&9y&)=&2(-1) \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{rrrrr} 10x&-&18y&=&2\\ -10x&+&18y&=&-2\\\cline{1-5} 0&+&0&=&0 \end{array}\qquad \impliedby \textit{another way of saying \underline{infinite solutions}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%20-5x%26%2B%269y%26%3D%26-1%20%5Cend%7Barray%7D~%5Chfill%20%5Cimplies%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsecond%20equation%20%7D%5Ctimes%202%7D%7B%20%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%202%28-5x%26%2B%269y%26%29%3D%262%28-1%29%20%5Cend%7Barray%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%20-10x%26%2B%2618y%26%3D%26-2%5C%5C%5Ccline%7B1-5%7D%200%26%2B%260%26%3D%260%20%5Cend%7Barray%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Banother%20way%20of%20saying%20%5Cunderline%7Binfinite%20solutions%7D%7D)
if we were to solve both equations for "y", we'd get

notice, the 1st equation is really the 2nd in disguise, since both lines are just pancaked on top of each other, every point in the lines is a solution or an intersection, and since both go to infinity, well, there you have it.