Someone please help me!!
1 answer:
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Answer:
<em>The coordinates of P are (-2,0)</em>
<em>The radius of the circle is 5.</em>
Step-by-step explanation:
<u>Analytic geometry</u>
The diagram shows a circle with center P, and two points A(-6,3) and B(2,-3) that form the diameter of the circle.
a)
The center of the circle lies at the midpoint of A and B. The midpoint (xm,ym) can be calculated by:
Substituting x1=-6, x2=2, y1=3, y2=-3:
Thus, the coordinates of P are (-2,0)
b) The radius of the circle is the distance from the center to any point in its circumference. We can use the distance from P to A or B indistinctly.
Given two points A(x1,y1) and P(x2,y2), the distance between them is:
Substituting x1=-6, x2=-2, y1=3, y2=0:
The radius of the circle is 5.
• First way to solve:
We'll manipulate the expression of the equation:
If we have y=0:
Then, the function has one real zero (x=3) and two imaginary zeros (4i and -4i).
Answer: B
• Second way to solve:
The degree of the function is 3. So, the function has 3 complex zeros.
Since the coefficients of the function are reals, the imaginary roots are in a even number (a imaginary number and its conjugated)
The function "has only one non-repeated x-intercept", then there is only one real zero.
The number of zeros is 3 and there is 1 real zero. So, there are 2 imaginary zeros.
Answer: B.
We'll manipulate the expression of the equation:
If we have y=0:
Then, the function has one real zero (x=3) and two imaginary zeros (4i and -4i).
Answer: B
• Second way to solve:
The degree of the function is 3. So, the function has 3 complex zeros.
Since the coefficients of the function are reals, the imaginary roots are in a even number (a imaginary number and its conjugated)
The function "has only one non-repeated x-intercept", then there is only one real zero.
The number of zeros is 3 and there is 1 real zero. So, there are 2 imaginary zeros.
Answer: B.