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
Let: eq 1:2x+5y=-55
eq 2:y=3x+6
by substituting eq 2 in eq 1 we get,
2x+5(3x+6)=-55
2x+15x+30+55=0
17x + 85=0
x=-85/17
x=-5
By substituting x value in eq 2,
we have,
y=3(-5)+6=-15+6=-9
Answer:
12.76
Step-by-step explanation:
This is much easier to solve if you draw it out on a graph. The perimeter is the sum of all the sides. I used the Pythagorean theorem to find the hypotenuse.


Answer:
18 and 19
Step-by-step explanation: