Given: AD ≅ BC and AD ∥ BC
Prove: ABCD is a parallelogram.
Statements Reasons
1. AD ≅ BC; AD ∥ BC 1. given
2. ∠CAD and ∠ACB are alternate interior ∠s 2. definition of alternate interior angles
3. ∠CAD ≅ ∠ACB 3. alternate interior angles are congruent
4. AC ≅ AC 4. reflexive property
5. △CAD ≅ △ACB 5. SAS congruency theorem
6. AB ≅ CD 6. Corresponding Parts of Congruent triangles are Congruent (CPCTC)
7. ABCD is a parallelogram 7. parallelogram side theorem
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Step-by-step explanation:

The simplest method is "brute force". Calculate each term and add them up.
∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)
∑ = 3 + 6 + 9 + 12 + 15
∑ = 45

∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²
∑ = 4 + 16 + 36 + 64
∑ = 120

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)
∑ = -4 + -2 + 0 + 2
∑ = -4
4. 1 + 1/4 + 1/16 + 1/64 + 1/256
This is a geometric sequence where the first term is 1 and the common ratio is 1/4. The nth term is:
a = 1 (1/4)ⁿ⁻¹
So the series is:

5. -5 + -1 + 3 + 7 + 11
This is an arithmetic sequence where the first term is -5 and the common difference is 4. The nth term is:
a = -5 + 4(n−1)
a = -5 + 4n − 4
a = 4n − 9
So the series is:

Answer:
c. 1 and 3
Step-by-step explanation:
To quickly solve this problem, we can use a graphing tool or a calculator to plot each equation.
Please see the attached image below, to find more information about the graph
s
The equations are:
1) y = sin (3x + π/6)
2) y = cos (3x - π/6)
3) y = cos (3x - π/3)
Looking at the graphs, we can see that the identical ones
are equations one and three
Correct option:
c. 1 and 3
We have to choose the correct answer for the center of the circumscribed circle of a triangle. The center of the circumscribed circle of a triangle is where the perpendicular bisectors of a triangle intersects. In this case P1P2 and Q1Q2 are perpendicular bisectors of sides AB and BC, respectively and they intersect at point P. S is the point where the angle bisectors intersect ( it is the center of the inscribed circle ). Answer: <span>P.</span>