Answer:
- 0.5 + 2.985i
- 1 + 2.828i
- 1.5 + 2.598i
- 2 + 2.236i
Explanation:
Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.
Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.
The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:
Then, the work consists in finding pairs (a,b) for which:
You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:
I will use b =0.5, b = 1, b = 1.5, b = 2
Then, four distinct complex numbers that have an absolute value of 3 are:
- 0.5 + 2.985i
- 1 + 2.828i
- 1.5 + 2.598i
- 2 + 2.236i
Answer:
9165.0
Step-by-step explanation:
13*94*7.5 = 9165.0
Answer:
104.8 in^2
Step-by-step explanation:
There are 2 ways to solve this problem.
The 1st way:
Let's make 2 triangles and 1 rectangle:
Rectangle Length = 8.3
Rectangle Width = 8
So, the left out length will be 17.9 - 8.3
=> 9.6
Since, 9.6 cm is for 2 parts.
=> 9.6 / 2
=> 4.8
So, Height of the Triangle = 8
Base of the triangle = 4.8
Area of a rectangle
=> 8.3 x 8
=> 66.4
Area of the triangle
=> 1/2 x 8 x 4.8
=> 4 x 4.8
=> 19.2
There are 2 triangles:
=> 19.2 x 2
=> 38.4
=> 66.4 + 38.4
=> 104.8
The area of the trapezoid = 104.8 in^2.
The 2nd way is:
Area of a trapezoid
=> Smaller Base + Larger Base / 2 x Height
=> 8.3 + 17.9 / 2 x 8
=> 26.2 / 2 x 8
=> 13.1 x 8
=> 104.8
The area of the trapezoid is 104.8 in^2
Answer:
C. (see the attachment)
Step-by-step explanation:
Both inequalities include the "or equal to" case, so both boundary lines will be solid. That excludes choices A and D.
The first inequality is plotted the same way in all graphs, so we must look at the second inequality. The relationship of y and the comparison symbol is ...
-y ≥ (something)
If we multiply by -1, we get ...
y ≤ (something else)
This means the solution space will be <em>on or below (less than or equal to) the boundary line</em>. This is the shaded area in graph C. (Graph B shows shading <em>above</em> the line.)
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<em>Further comment</em>
Since the boundary for the second inequality is fairly steep, "above" and "below" the line can be difficult to see. Rather, you can consider the relationship of x to the comparison symbol. For the second inequality, that is ...
x ≥ (something)
indicating the solution space is <em>on or to the right of the boundary line</em>.
