You never put in what the question was.
Slope formula: ![\frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
Slope of line one: ![-\frac{1}{2}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D)
Slope of line two: ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
Slope of line three: ![-\frac{2}{3}](https://tex.z-dn.net/?f=-%5Cfrac%7B2%7D%7B3%7D)
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Answer:
Down below hope it helps :)
Step-by-step explanation:
The evidence "SAY" was only written once. It wasn't by the same article, the writer introduces two separate articles.
Kinda confused by this question but I think that's the answer.
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
Square roots and squaring are always nonlinear.