Find the greatest common factor: 4ab^2 c^4-2a^2 b^3 c^2+6a^3 b^4 c
1 answer:
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Answer: See step by step
Step-by-step explanation: For A my 15 statements are.
- It has 3 triangles inside it, ACD, ADC, and ABC
- It has 2 right triangles, and 1 isoceles
- AC≅AB
- CD≅DB
- D is midpoint of CB
- AD⊥CB
- Angle CDA=90 degrees
- Angle BDA equal 90 degrees
- AD≅AD
- ΔCDA≅ΔBDA by any congruence theorem, (SSS, SSA,AAS,ASA, HL)
-
+
=
12.+
=
13. Triangle ABC has a max of 180 degrees.
14. We can rotate this triangle 180 degrees and it will coincide.
15. We can reflect triangle ACD over vertical line ACD and it will be congruent to ABD.
2. We use pythagorean theorem since it has a right angle.
+
=
Let plug it in.
+
=
1600+ b^2=2025
b^2=424
sqr root of 425 is about 21. Now let find the perimeter.
AB is 45, Since BD+DC=CB, and they are congruent they are equal so 21+21=42 and AC is congruent to AB so it is 45. So the perimeter is 132.
For 3. Start at the orgin, then go up 5 on the y-axis so you should be at (0,5)
Then use the rise over run method to graph it. go left -3 and and up 1. Keep doing that 2 more times then draw a straight line.
Your 3 point should be
(0,5)
(1,2)
(2,1)
Answer:
Step-by-step explanation:
1). In ΔABC,
For angle A as a reference angle,
Opposite side of ∠A → BC
Adjacent side of ∠A → AC
Hypotenuse → AB
2). By using tangent ratio in ΔYWZ,
tan(72°) =
tan(72°) =
x =
x = 3.899
x ≈ 3.9
3). By using sine rule in ΔKNM,
tan(x°) =
tan(x) =
x =
x = 19.03
x ≈ 19.0°
4). By applying cosine ratio in ΔJKF,
tan(70°) =
tan(70) =
x =
x = 21.838
x ≈ 21.8