Answer:
∠B ≅ ∠F ⇒ proved down
Step-by-step explanation:
<em>In the </em><em>two right triangles</em><em>, if the </em><em>hypotenuse and leg</em><em> of the </em><em>1st right Δ ≅</em><em> the </em><em>hypotenuse and leg</em><em> of the </em><em>2nd right Δ</em><em>, then the </em><em>two triangles are congruent</em>
Let us use this fact to solve the question
→ In Δs BCD and FED
∵ ∠C and ∠E are right angles
∴ Δs BCD and FED are right triangles ⇒ (1)
∵ D is the mid-point of CE
→ That means point D divides CE into 2 equal parts CD and ED
∴ CD = ED ⇒ (2) legs
∵ BD and DF are the opposite sides to the right angles
∴ BD and DF are the hypotenuses of the triangles
∵ BD ≅ FD ⇒ (3) hypotenuses
→ From (1), (2), (3), and the fact above
∴ Δ BCD ≅ ΔFED ⇒ by HL postulate of congruency
→ As a result of congruency
∴ BC ≅ FE
∴ ∠BDC ≅ ∠FDE
∴ ∠B ≅ ∠F ⇒ proved
Answer:
-2f+10+11f
= 9f+10
Step-by-step explanation:
just expand it
Y = 5x + 3
when x = 4
y = 5(4) + 3 = 23
Answer (4,23)
Answer:
I think that the answer would be C because you have to multiply the term at the top. This would mean you would not do C. Sorry if this is wrong. At least you have your eliminating process now. lol so doing C is out of the box, which makes C your answer.
Step-by-step explanation:
Answer:
proof is below
Step-by-step explanation:
Answer:
25^9 + 5^17=
=2^18 + 5^17
=5^17(5+1)
=5^17×6
=5^16×30 is divisible by 30