Answer:
ASA
ΔFGH ≅ ΔIHG ⇒ answer B
Step-by-step explanation:
* Lets revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and
including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ
≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles
and one side in the 2ndΔ
- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse
leg of the 2nd right angle Δ
* Lets prove the two triangles FGH and IHG are congruent by on of
the cases above
∵ FG // HI and GH is transversal
∴ m∠FGH = m∠IHG ⇒ alternate angles
- In the two triangles FGH and IHG
∵ m∠FHG = m∠IGH ⇒ given
∵ m∠FGH = m∠IHG ⇒ proved
∵ GH = HG ⇒ common side
∴ ΔFGH ≅ ΔIHG ⇒ ASA
* ASA
ΔFGH ≅ ΔIHG
Answer:
1.A
2.D
Step-by-step explanation:
Brainliest pls hardstuck ambitious
Answer:
Scalene acute
Step-by-step explanation:
All angles measures are less than 90 so it would be acute. No sides are equal so it would be scalene
Answer:
Answer is B - 290.
Step-by-step explanation:
The idea here is that we're assuming the frequency each colour is drawn is a reflection of the proportion of marbles in the bag that are that colour. The more of a particular colour you have, the more likely you are to draw it, and vice versa.
Therefore we can assume the proportion of blue, gree, and red marbles is 13%, 29%, and 58% respectively based on the first 100 draws.
These percents should stay relatively similar as more draws are done, approaching the true proportions of each colour better the higher the number of draws. That means 29%, or 290, of the 1000 draws will be green.