Find a degree 3 polynomial with real coefficients having zeros 2 and 2 − 2 i and a lead coefficient of 1. Write P in expanded fo
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The statements that are true about the intervals of the continuous function are Options 2, 4 and 5
- f(x) ≤ 0 over the interval [0, 2].
- f(x) > 0 over the interval (–2, 0).
- f(x) ≥ 0 over the interval [2, ).
Looking at the values given, the intervals which satisfies the condition are known to be:
f(x)<=0 over the interval [0,2]
f(x)>0 over the interval (-2,0)
f(x)>=0 over the interval [2,∞)
Because:
Since the table with x and f(x) values, we have to examine analyze the table and see the each option that is in line with f(x) or not .
Examine the values of x that is from -3 to 3, the f(x) values are both positive and negative . hence f(x)>0 is false over the interval (-∞,3)
Looking at the the interval from 0 to 2, the f(x) values are 0 and negative. Hence, f(x)<=0 over the interval [0,2]
When you look over the interval (-1,1), the f(x) values are said to be both positive and negative and as such, f(x)<0 is false over the interval (-1,1)
When you look at the interval (-2,0) , the f(x) is positive and as such, f(x)>0 over the interval (-2,0)
Looking at the interval [2,∞), f(x) is positive and as such, f(x)>=0 over the interval [2,∞)
Therefore, Option 2, 4 and 5 are correct.
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Answer:
Step-by-step explanation:
Given
Required
Determine the height
The volume is calculated as:
Since, it has a triangular cross-section, the expression becomes:
So, we have:
The blue point on the number line is 5 units away from zero, which is: 5.
<h3>What is a Number Line?</h3>A number line is an arrangement of numbers from point zero, on both sides of the starting point, which is zero.
In the number line given, the blue point is 5 units away from the starting point, zero.
Therefore, the blue point is 5 on the number line.
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