A rotation 270° counterclockwise about the origin is the same as rotation 90° clockwise about the origin and has a rule:
(x,y)→(y,-x).
Then:
- D(−2,4)→D'(4,2)
- E(4,7)→E'(7,-4)
- F(10,3)→F'(3,-10)
- G(8,0)→G'(0,-8)
Answer: the coordinates of vertices of quadrilateral D′E′F′G′ are D'(4,2), E'(7,-4), F'(3,-10), G'(0,-8).
Answer:
3x2=6
Step-by-step explanation:
steven has three times as much as sahrah so she is going to have 6 stamps
Answer:
C. 54π + 20.25√3 cm²
Step-by-step explanation:
The shaded area can be split into two areas: a sector and an isosceles triangle.
Area of a sector is:
A = (θ/360°) πr²
where θ is the central angle and r is the radius.
Area of an isosceles triangle can be found with SAS formula:
A = ½ ab sin θ
where a and b are two sides of a triangle and θ is the angle between them.
In this case, r = a = b = 9 cm. The central angle of the sector is 240°, and the vertex angle of the triangle is 120°. Therefore, the total area is:
A = (240°/360°) π (9 cm)² + ½ (9 cm) (9 cm) sin 120°
A = 54π + 20.25√3 cm²
The slope for that is undefined.
Answer:
A. Repeat the simulation several more times
Step-by-step explanation:
The purpose of the simulation model is to represent the effectivity of the passes.
The proportion of successful passes is 60%.
As we have 10 digits available, 6 (digits from 0 to 5) are used for the outcome "the pass is completed" and 4 (digits 6, 7, 8, and 9) to represent the outcome "the pass is not completed". This is correct, as it represents a probability of 60% of having a successful pass.
But to have a representative distribution of the possible and probable results, the simulation have to run enough times to have a stable distribution of the results.
