Solve the inequality –24 ≤ x – 3 – 8x. A. x ≥ –3 B. x ≥ 3 C. x ≤ 3 D. x ≥

Can someone help me do part two please? It’s very important send a picture or something. I don’t even care if you tell me the st

Nataly_w [17]

<h3>Explanation:</h3>

1. "Create your own circle on a complex plane."

The equation of a circle in the complex plane can be written a number of ways. For center c (a complex number) and radius r (a positive real number), one formula is ...

|z-c| = r

If we let c = 2+i and r = 5, the equation becomes ...

|z -(2+i)| = 5

For z = x + yi and |z| = √(x² +y²), this equation is equivalent to the Cartesian coordinate equation ...

(x -2)² +(y -1)² = 5²

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2. "Choose two end points of a diameter to prove the diameter and radius of the circle."

We don't know what "prove the diameter and radius" means. We can show that the chosen end points z₁ and z₂ are 10 units apart, and their midpoint is the center of the circle c.

For the end points of a diameter, we choose ...

z₁ = 5 +5i
z₂ = -1 -3i

The distance between these is ...

|z₂ -z₁| = |(-1-5) +(-3-5)i| = |-6 -8i|

= √((-6)² +(-8)²) = √100

|z₂ -z₁| = 10 . . . . . . the diameter of a circle of radius 5

The midpoint of these two point should be the center of the circle.

(z₁ +z₂)/2 = ((5 -1) +(5 -3)i)/2 = (4 +2i)/2 = 2 +i

(z₁ +z₂)/2 = c . . . . . the center of the circle is the midpoint of the diameter

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3. "Show how to determine the center of the circle."

As with any circle, the center is the <em>midpoint of any diameter</em> (demonstrated in question 2). It is also the point of intersection of the perpendicular bisectors of any chords, and it is equidistant from any points on the circle.

Any of these relations can be used to find the circle center, depending on the information you start with.

As an example. we can choose another point we know to be on the circle:

z₄ = 6-2i

Using this point and the z₁ and z₂ above, we can write three equations in the "unknown" circle center (a +bi):

|z₁ - (a+bi)| = r
|z₂ - (a+bi)| = r
|z₄ - (a+bi)| = r

Using the formula for the square of the magnitude of a complex number, this becomes ...

(5-a)² +(5-b)² = r² = 25 -10a +a² +25 -10b +b²

(-1-a)² +(-3-b)² = r² = 1 +2a +a² +9 +6b +b²

(6-a)² +(-2-b)² = r² = 36 -12a +a² +4 +4b +b²

Subtracting the first two equations from the third gives two linear equations in a and b:

11 -2a -21 +14b = 0

35 -14a -5 -2b = 0

Rearranging these to standard form, we get

a -7b = -5

7a +b = 15

Solving these by your favorite method gives ...

a +bi = 2 +i = c . . . . the center of the circle

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4. "Choose two points, one on the circle and the other not on the circle. Show, mathematically, how to determine whether or not the point is on the circle."

The points we choose are ...