Answer:
2.5
Step-by-step explanation:
A triangle ABC with vertices A(0, 0), B(0, 4), and C(6, 0) is dilated to form a triangle A'B'C' with vertices A′(0, 0), B′(0, 10), and C′(15, 0). If the center of dilation is point A or A' (the origin), then
Since
- AB=4;
- A'B'=10;
- AC=6;
- A'C'=15,
then the scale factor of the dilation is
I believe the sum of the root is: -3/2
And the product of the root is: 0
I hope this helps and have a great week :)
Answer:
Step-by-step explanation:
A1. C = 104°, b = 16, c = 25
Law of Sines: B = arcsin[b·sinC/c} ≅ 38.4°
A = 180-C-B = 37.6°
Law of Sines: a = c·sinA/sinC ≅ 15.7
A2. B = 56°, b = 17, c = 14
Law of Sines: C = arcsin[c·sinB/b] ≅43.1°
A = 180-B-C = 80.9°
Law of Sines: a = b·sinA/sinB ≅ 20.2
B1. B = 116°, a = 11, c = 15
Law of Cosines: b = √(a² + c² - 2ac·cosB) = 22.2
A = arccos{(b²+c²-a²)/(2bc) ≅26.5°
C = 180-A-B = 37.5°
B2. a=18, b=29, c=30
Law of Cosines: A = arccos{(b²+c²-a²)/(2bc) ≅ 35.5°
Law of Cosines: B = arccos[(a²+c²-b²)/(2ac) = 69.2°
C = 180-A-B = 75.3°
We have an inequation:
x/3 < 3
⇒ x < 3*3
⇒ x < 9
The final answer is x < 9~
