To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Mark points (0,0), (-1,-1), (1,-1), (-2,-4), (2,-4)
Answer:
the answer is 612 -4n4+6n-5772
Answer:
I think the answer is A
Step-by-step explanation:
So sorry if the answer is incorrect
Answer:
The correct option is;
B
Step-by-step explanation:
The given system of inequalities are;
5·x - 4·y > 4...(1)
x + y < 2...(2)
Representing both inequalities as a function of "y", gives;
For, 5·x - 4·y > 4...(1), we have;
-4·y > 4 - 5·x
y < 4/(-4) - 5·x/(-4)
∴ y < 5·x/4 - 1
For x + y < 2...(2), we have;
y < 2 - x
Therefore, y is less than the values given by the equation of the straight line equalities, and the feasible region is given by the common region under both dashed lines representing both inequalities as shown in the attached diagram created using Microsoft Excel
The correct option is therefore, B.