Triangle RTS is congruent to RQS by AAS postulate of congruent
Step-by-step explanation:
Let us 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 1st Δ ≅ 2 angles
and one side in the 2nd Δ
- HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ
∵ SR bisects angle TSQ ⇒ given
∴ ∠TSR ≅ ∠QSR
∴ m∠TSR ≅ m∠QSR
∵ ∠T ≅ ∠Q ⇒ given
∴ m∠T ≅ m∠Q
In two triangles RTS and RQS
∵ m∠T ≅ m∠Q
∵ m∠TSR ≅ m∠QSR
∵ RS is a common side in the two triangle
- By using the 4th case above
∴ Δ RTS ≅ ΔRQS ⇒ AAS postulate
Triangle RTS is congruent to RQS by AAS postulate of congruent
Learn more:
You can learn more about the congruent in brainly.com/question/3202836
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Answer:
These lines are perpendicular.
Step-by-step explanation:
Put both equations in the slope intercept form of a line.
y = -6x - 8 This is already in the slope intercept form for a line
-x + 6y = 12 Add x to both sides of the equation
6y = x + 12 Divide through the whole equation by 6
y = 1/6x +2
Now we compare the two slopes of -6 and 1/6. They are negative reciprocals of each other. That means that the line are perpendicular.
Answer:
The answer to your question is: ![\sqrt[4]{2^{3} } + 1](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B2%5E%7B3%7D%20%7D%20%2B%201)
Step-by-step explanation:
= ![\sqrt[4]{2^{3} } + 1](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B2%5E%7B3%7D%20%7D%20%2B%201)
Answer:
I think the answer to the question is 5x
Cylinder has both circles top and bottom.