If a polynomial "contains", in a multiplicative sense, a factor
, then the polynomial has a zero at
.
So, you polynomial must contain at least the following:

If you multiply them all, you get

Now, if you want the polynomial to be zero only and exactly at the four points you've given, you can choose every polynomial that is a multiple (numerically speaking) of this one. For example, you can multiply it by 2, 3, or -14.
If you want the polynomial to be zero at least at the four points you've given, you can multiply the given polynomial by every other function.
If in the first quarter they scored 1/6 of their points, this would mean in the first quarter they scored 6 points (36/6=6). If in the second quarter they scored 1/4 of their points, they scored 9 points (36/4=9). If in the third quarter they scored 1/3 of their points, they scored 12 points (36/3=12).
12+9+6=27.
36 - 27 = 9
the answer is c, 9
Answer:
b=c-ax/y
Step-by-step explanation:
1) ax + by = c
-ax
2) by = c - ax
3) by/y = b and c - ax/y
4) b = c-ax/y
Answer:
y = 1/2 x + 3
Step-by-step explanation:
mid point of line BC = (2, 4)
slope of the line containing median (m) = 1/2
y intercept (b)= 3
since y = mx + b
y = 1/2x + 3