10m^3-m^2-6m+9 is the answer
Answer:
(2, 1)
Step-by-step explanation:
The best way to do this to avoid tedious fractions is to use the addition method (sometimes called the elimination method). We will work to eliminate one of the variables. Since the y values are smaller, let's work to get rid of those. That means we have to have a positive and a negative of the same number so they cancel each other out. We have a 2y and a 3y. The LCM of those numbers is 6, so we will multiply the first equation by a 3 and the second one by a 2. BUT they have to cancel out, so one of those multipliers will have to be negative. I made the 2 negative. Multiplying in the 3 and the -2:
3(-9x + 2y = -16)--> -27x + 6y = -48
-2(19x + 3y = 41)--> -38x - 6y = -82
Now you can see that the 6y and the -6y cancel each other out, leaving us to do the addition of what's left:
-65x = -130 so
x = 2
Now we will go back to either one of the original equations and sub in a 2 for x to solve for y:
19(2) + 3y = 41 so
38 + 3y = 41 and
3y = 3. Therefore,
y = 1
The solution set then is (2, 1)
Answer:
control group
Step-by-step explanation:
Answer:
Step-by-step explanation:
Let us assume that if possible in the group of 101, each had a different number of friends. Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.
Say P has 100 friends then P has all other persons as friends. In this case, there cannot be any one who has 0 friend. So a contradiction. Hence proved
Part ii: EVen if instead of 101, say n people are there same proof follows as
if different number of friends then they would be 0,1,2...n-1
If one person has n-1 friends then there cannot be any one who does not have any friend.
Thus same proof follows.
Answer:
3’d
Step-by-step explanation:
