Select all of the points that are solutions to the system of linear inequalities that is listed below 10x + 4y < 12 8x -3y &g
1 answer:
Answer:
(3, -8) and (2, -10)
Step-by-step explanation:
Given
Required
Select the true coordinate points in (3, -8) (2, 5) (-5, 1) (10, 3) (2, -10)
(3, -8)
x = 3 and y = -8
--- True
--- True
(2, 5)
x = 2 and y = 5
--- False (No need to check the other inequality)
(-5, 1)
x = -5 and y = 1
--- True
--- False
(10, 3)
x = 10 and y = 3
--- False (No need to check the other inequality)
(2, -10)
x = 2 and y = -10
--- True
--- True
Hence, the solution to the inequalities are:
(3, -8) and (2, -10)
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Answer:
To solve a polynomial inequality, we factor the polynomial into irreducible factors and find all the real _zeros_ polynomial. Then we find the intervals determined by the real _zeros and use test points in each interval to find the_ sign of the polynomial on that interval.
If P(x) = x(x+2)(x-1)
And P(x) ≥ 0
We see that P(x) ≥ 0 on the intervals (-2, 0) and (1, ∞).
Step-by-step explanation:
The complete question is attached to this solution
To solve inequality of a polynomial, we first obtain the solutions of the polynomial. The solutions of the polynomial are called the zeros of the polynomial.
If P(x) = x(x+2)(x-1)
The solutions of this polynomial, that is the zeros of this polynomial are 0, -2 and 1.
To now solve the inequality that arises when
P(x) ≥ 0
We redraw the table and examine the intervals
The intervals to be examined as obtained from the zeros include (-∞, -2), (-2, 0), (0, 1) and (1, ∞)
Sign of | x<-2 | -2<x<0 | 0<x<1 | x>1
x | -ve | -ve | +ve | +ve
(x + 2) | -ve | +ve | +ve | +ve
(x - 1) | -ve | -ve | -ve | +ve
x(x+2)(x-1) | -ve | +ve | -ve | +ve
The intervals that satisfy the polynomial inequality P(x) = x(x+2)(x-1) ≥ 0 include
(-2, 0) and (1, ∞)
Hope this Helps!!!
