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:A
Step-by-step explanation:
Given
Turning Point is the point where the graph changes its nature i.e. either increasing to decreasing or decreasing to increasing
to find turning point
For critical point
F'(x)=0
x+2=0
x=-2
thus
(-2,-4 ) is the turning Point where Function changes its nature
The angles in an equilateral triangle are 60 degrees.
sin 60 = 9 / h where h is hypotenuse ( = side of the triangle)
h = 9 / sin 60 = 10.392
Perimeter = 3 * h = 31.18 to nearest hundredth
Answer:
What is the expresion I am confused?
Step-by-step explanation:
