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:
2x^2x(x^+3x-2)
Step-by-step explanation:
Answer:
A. x = 3.03
B. x = 4/3
Step-by-step explanation:
A. 2000 (x-0.03) = 6000 is quickly simplified byu dividing both sides by 2000:
(x - 0.03) = 3. Removing the parentheses, we get: x = -0.03 = 3, or x = 3.03.
B. The distributive property is the faster method here. We determine that the LCD is 12 and multiply both sides of this equation 1/4 (4+x) = 4/3 by 12:
3(4 + x) = 16
and then carry out the indicated multiplication: 12 + 3x = 16, or
3x = 4, or x = 4/3
Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.
Exact Form:
Decimal Form:
2701.99962990
Thus, <em>2,701</em> is your answer
