I think it is negative five
The sector (shaded segment + triangle) makes up 1/3 of the circle (which is evident from the fact that the labeled arc measures 120° and a full circle measures 360°). The circle has radius 96 cm, so its total area is π (96 cm)² = 9216π cm². The area of the sector is then 1/3 • 9216π cm² = 3072π cm².
The triangle is isosceles since two of its legs coincide with the radius of the circle, and the angle between these sides measures 120°, same as the arc it subtends. If b is the length of the third side in the triangle, then by the law of cosines
b² = 2 • (96 cm)² - 2 (96 cm)² cos(120°) ⇒ b = 96√3 cm
Call b the base of this triangle.
The vertex angle is 120°, so the other two angles have measure θ such that
120° + 2θ = 180°
since the interior angles of any triangle sum to 180°. Solve for θ :
2θ = 60°
θ = 30°
Draw an altitude for the triangle that connects the vertex to the base. This cuts the triangle into two smaller right triangles. Let h be the height of all these triangles. Using some trig, we find
tan(30°) = h / (b/2) ⇒ h = 48 cm
Then the area of the triangle is
1/2 bh = 1/2 • (96√3 cm) • (48 cm) = 2304√3 cm²
and the area of the shaded segment is the difference between the area of the sector and the area of the triangle:
3072π cm² - 2304√3 cm² ≈ 5660.3 cm²
49.57(total amount paid) -2.5 (del fee) = 47.07
47.07/3= 15.69 is the price of each
Answer:
63 % for cats and 37 % for dogs
Step-by-step explanation:
Hey fellow Army.
So because JKLM=PQRS, that means that all the line segments are equal.
JK=PQ=6.4
KL=QR=13
LM=RS=9
MJ=SP=18
Okay now for the graph.
It was first reflected over the y axis, and then shifted over vertically by 9 .
Hope this helps. Have a good day!
