Answer:
Area of shaded region = 16π in² (D)
Step-by-step explanation:
The question is incomplete without the diagram if the circles. Find attached the diagram used in solving the question.
Area of the smaller circle = 8π in²
Area of a circle = πr²
πr² = 8π
r² =8
r = √8 = 2√2
From the diagram, there are two smaller circles in a bigger circle.
The radius of the bigger circle (R) is 2times the radius of the smaller circle (r)
R = 2r
Area of bigger circle = πR²
= π×(2r)² = π×(2×2√2)²
= π×(4√2)² = π×16×(√2)²
Area of bigger circle = π×16×2
Area of bigger circle = 32π in²
Since there are two smaller circles in a bigger circle
Area of shaded region = Area of bigger circle -2(area of smaller circles)
Area of shaded region = 32π in² - 2(8π in²)
Area of shaded region = 32π in² - 16π in²
Area of shaded region = 16π in²
X = the number of laps Joe has ran.
x + x + x + 6 = 222
3x = 216
x = 72
Jimmy ran 72 laps.
Joe ran 72 laps.
Sally ran 78 laps.
Answer:
IDK i am trying to solve the same problem. I am sorry
Step-by-step explanation:
Step-by-step explanation:

Answer:
B:
4 units right and six units up