Answer:
The values of k will be:
Step-by-step explanation:
Let the expression of polynomial P be
Let the expression if the polynomial Q be
Plug in Q(x) = 0
0 = x+2
x = -2
As (x+2) is a factor of 3x²-4kx-4k²
substitute x = -2 in the the polynomial
3x²-4kx-4k² = 0
Write in the standard form ax²+bx+c = 0
Factor out common term -4
Factor k²-2k-3: (k+1)(k-3)
Using the zero factor principle
if ab=0, then a=0 or b=0 (or both a=0 and b=0)
solving k+1=0
k+1 = 0
k = -1
solving k-3=0
k-3=0
k = 3
Thus, the values of k will be:
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Answer: (-4,-6) is the point that ALMOST satisfies both inequalities. IF they were equalities, this would be the solution.
The question is a bit confusing as it asks for "which points (x,y) satisfies both" It's ungrammatical, and many points (infinite within the shaded region) are solutions that SATISFY the system of inequalities!
Step-by-step explanation: Substitute the x and y-values and see if the inequalities are true.
y>x-2 -6> -4-2 -6= -6
That point (-4,-6) is on the dashed line, so not exactly a true solution; this is a question about inequalities. So y values have to be greater than-6 or x-values less than -4 for a true inequality.
y>2x+2
-6>(2)(-4) +2
-6> -8 +2
-6> -6 Again, equal, so for this y-values have to be greater than-6 and/or x-values less than -4 in order to have a true inequality.
If you have the graph to look at, you can select any points in the shaded region that satisfies both of the inequalities.
The answer two your problem is no remainder again because 46 divided by 2 equals 23
