Can you post the image because I need to clarify in your trigonometry hw
Answer:
See Explanation
Step-by-step explanation:
The question is incomplete, as Hernando and Rachel's solution are not provided. So, I will just solve the question directly.
Given
Required
Factor
Group into 2
![2mp-6p+27-9m = [2mp-6p]+[27-9m]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%20%5B2mp-6p%5D%2B%5B27-9m%5D)
Factor each group
![2mp-6p+27-9m = 2p[m-3]+9[3-m]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%202p%5Bm-3%5D%2B9%5B3-m%5D)
Rewrite 3 - m as -(m-3)
So, we have:
![2mp-6p+27-9m = 2p[m-3]-9[m-3]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%202p%5Bm-3%5D-9%5Bm-3%5D)
Factor out m - 3
![2mp-6p+27-9m = [2p-9][m-3]](https://tex.z-dn.net/?f=2mp-6p%2B27-9m%20%3D%20%5B2p-9%5D%5Bm-3%5D)
5/9
Using a table of outcomes, you can see that 20 of the possible 36 outcomes are even totals. Simplify for the 5/9 or roughly 56%
If you divide 9/2 you receive 4.5. Tho when you divide 1/3 you get .33333 (repeated), so now you know that both the fractions do not correlate. Now the 'multiples' of 1/3 are 2/ 6, 3/9, 4/12, 5/15, 6/18, 7/21 and so on. If you do the 'multiples' of 2/9 you get 4/18, 6/27. The common factors are 4/18 and 6/18. So know you know that Sandy is using 2/18 (or 1/9) cups more of sugar.
