I'm not sure you are asking the right question? it tells you what the radius is right there. What is the question asking
Answer: C
Step-by-step explanation:
You can solve this by first multiplying both sides of the fraction by the conjugate of the denominator.
In this case, the conjugate is (8 - 2i).
So:
Simplify:
Simplify again, knowing that i^2 = -1
Then, divide both numbers in the numerator by 68 to get your answer.
Answer:
Step-by-step explanation:
Let
x------> the cost of each ticket
we know that
----> inequality that represent the situation
solve for x
Divide by 3 both sides
This pattern of question is always coming up. Since we can't easily guess, then let us set up simultaneous equation for the statements.
let the two numbers be x and y.
Multiply to 44. x*y = 44 ..........(a)
Add up to 12. x + y = 12 .........(b)
From (b)
y = 12 - x .......(c)
Substitute (c) into (a)
x*y = 44
x*(12 - x) = 44
12x - x² = 44
-x² + 12x = 44
-x² + 12x - 44 = 0.
Multiply both sides by -1
-1(-x² + 12x - 44) = -1*0
x² - 12x + 44 = 0.
This does not look factorizable, so let us just use quadratic formula
comparing to ax² + bx + c = 0, x² - 12x + 44 = 0, a = 1, b = -12, c = 44
x = (-b + √(b² - 4ac)) /2a or (-b - √(b² - 4ac)) /2a
x = (-(-12) + √((-12)² - 4*1*44) )/ (2*1)
x = (12 + √(144 - 176) )/ 2
x = (12 + √-32 )/ 2
√-32 = √(-1 *32) = √-1 * √32 = i * √(16 *2) = i*√16 *√2 = i*4*√2 = 4i√2
Where i is a complex number. Note the equation has two values. We shall include the second, that has negative sign before the square root.
x = (12 + √-32 )/ 2 or (12 - √-32 )/ 2
x = (12 + 4i√2 )/ 2 (12 - 4i√2 )/ 2
x = 12/2 + (4i√2)/2 12/2 - (4i√2)/2
x = 6 + 2i√2 or 6 - 2i√2
Recall equation (c):
y = 12 - x, When x = 6 + 2i√2, y = 12 - (6 + 2i√2) = 12 - 6 - 2i√2 = 6 - 2i√2
When x = 6 - 2i√2, y = 12 - (6 - 2i√2) = 12 - 6 + 2i√2 = 6 + 2i√2
x = 6 + 2i√2, y = 6 - 2i√2
x = 6 - 2i√2, y = 6 + 2i√2
Therefore the two numbers that multiply to 44 and add up to 12 are:
6 + 2i√2 and 6 - 2i√2
Answer:
A'
Step-by-step explanation:
It is A' because they are corresponding angles, it is the same exact shape it has just been translated to a different area of the graphic plane.
If my answer is incorrect, please notify me and I'll assess what it was that I did wrong.
