What is the sum? StartFraction 2 Over x squared EndFraction + StartFraction 4 Over x squared EndFraction StartFraction 6 Over x
squared EndFraction StartFraction 6 Over x Superscript 4 EndFraction StartFraction 6 Over 2 x squared EndFraction StartFraction 8 Over x squared EndFraction
2 answers:
Answer:
StartFraction 6 Over x squared EndFraction
Step-by-step explanation:
Obviously, you prefer words to math symbols, so we'll show the solution using words.
The sum StartFraction 2 Over x squared EndFraction + StartFraction 4 Over x squared EndFraction equals StartFraction 2 plus 4 Over x squared EndFraction equals StartFraction 6 Over x squared EndFraction
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In appropriate symbols this would be ...
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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!!!
Answer:
- 2
Step-by-step explanation:
Rate of change is the measure of the slope m
Calculate the slope using the slope formula
m =
with (x₁, y₁ ) = (0, 2 ) and (x₂, y₂ ) = (1, 0) ← 2 points on the line
m = =
= - 2