Answer:
(-4, 0) U (1, ∞)
Step-by-step explanation:
Set each factor EQUAL to zero to find the zeroes (since it is not actually equal to zero, you will use an open circle when graphing and an open bracket when writing in interval notation).
x = 0 x-1 = 0 x + 4 = 0
x = 1 x = -4
Next, choose a value to the far left, between each of the zeroes, and to the far right to evaluate if it makes a true statement when input into the given inequality.
far left (I choose -5): -5(-5 - 1)(-5 + 4) > 0 → (-)(-)(-) > 0 → negative > 0 FALSE
- 4 to 0 (I choose -2): -2(-2 - 1)(-2 + 4) > 0 → (-)(-)(+) > 0 → positive > 0 TRUE
0 to 1 (I choose 0.5): .5(.5 - 1)(.5 + 4) > 0 → (+)(-)(+) > 0 → negative > 0 FALSE
far right (I choose 2): 2(2 - 1)(2 + 4) > 0 → (+)(+)(+) > 0 → positive > 0 TRUE
Answer:
Polymomial with 4 terms i believe
Graph the feasible region for the following constraints: x + y < 5. 2x + y > 4 ... y = 10/3. x = 30/3 - 10/3 = 20/3. Intersects at (20/3, 10/3). -x + 2y = 0. x - 2y = 0.
hence i would go for
<span>(6,3)</span>
Answer:
RS = 5
Step-by-step explanation:
Since R is between Q and S , then
QS = QR + RS , that is
6x - 3 = 3x - 2 + 2x + 1
6x - 3 = 5x - 1 ( subtract 5x from both sides )
x - 3 = - 1 ( add 3 to both sides )
x = 2
Then
RS = 2x + 1 = 2(2) + 1 = 4 + 1 = 5
