The answer is 60.
You must first solve what is in parentheses.
3 (20)
60
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:
Step-by-step explanation:
4 lines of symmetry
A definitely has to be one because it's a fact and you need to know what it looks like. B isn't true because the greatest integer function rounds down values. C is also not correct because a one-to-one graph has one x-value for every y-value. This one doesn't. D has to be true because f[0] = 0 because the Greatest Integer Function requires it to be. And E is also correct because that's in the definition of a Greatest Integer Function. So, your answers are A, D and E.
